Ray Optics And Optical Instruments - Class 12 Engineering Physics - Extra Questions
Why does a Stick immersed in water appear bent and short? Explain with the help of a ray diagram.
Stick immersed in water appears bent due to the refraction of light at the air-water interface.
When rays from $$O$$ gets refracted at the interface at $$C$$ and $$D$$, it appears to come from $$I$$ So image formed is at $$I$$.
Match the entries of column I with entries of column II
Draw a schematic diagram of a reflecting telescope (Cassegrain). Write its two advantages over a refracting telescope.
Reflecting telescopes have many advantages over refracting telescopes.
Mirrors don't cause chromatic aberration and they are easier and cheaper to build large. They are also easier to mount because the back of the mirror can be used to attach to the mount.
Mirrors don't cause chromatic aberration and they are easier and cheaper to build large. They are also easier to mount because the back of the mirror can be used to attach to the mount.
How does the shift depend upon the thickness AD or BC of the block ?
Why does a beam of light bend when it enters glass at an angle ? Why does it not bend if it enters the glass at right angles ?
Draw and complete the following diagrams to show that what happens to the beams of light as they enter the glass block and then leave it :
An object is placed at a long distance in front of a convex mirror of radius of curvature 30 cm. State the position of its image.
(a) Give two circumstances in which a concave mirror can form a magnified image of an object placed in front of it. Illustrate your answer by drawing labelled ray diagrams for both.
(b) Which one of these circumstances enables a concave mirror to be used as a shaving mirror?
How does phenomenon of lateral inversion occurs
Why does white light split up into different colours while passing through a glass prism
Which is a better reflector- a plane mirror or a right - angled p rism
Explain the term critical angle with the aid of a labelled diagram.
Critical angle: The angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is $$90^{0}$$ is called the critical angle.
Show at least two rays emerging from the surface PQ after reflection from the surface RS.
AO is incident ray.
OC is refracted ray
D,E are two emergent rays.
How is the reflection of light ray from a plane mirror different from the refraction of light ray as it enters a block of glass ?
Trace the path of the ray of light through a glass slab, when it passes from water to a glass slab and what will happen to the incident ray ? Draw the ray diagram.
If both pieces of glass have the same refractive index and have parallel surfaces$$,$$ the rays in and out will be parallel$$.$$
An object is seen first in red light and then in violet light through a simple microscope. In which case is the magnification larger?
With a neat labelled diagram, explain the formation of image in a simple microscope.
AB is the size of object, A'B' is the size of image, F and F'is the focal point of lens. Arrows describe the direction of light ray. X and Y leads to the eyes of the observer
Draw the diagram given below and clearly show the path taken by the emergent ray
The required diagram is shown in the figure.
Define magnifying power of a simple microscope. How can it be increased?
Give reason:
The bottom of water tank appears raised.
What is magnifying glass? State its two uses.
If the magnification of a body of size 1 m is 2, what is the size of the image?
What is refraction? Why pencil appears bent in water?
Draw a labelled ray diagram of a reflecting telescope. Mention its two advantages over the refracting telescope.
Advantages of reflecting telescope over a refracting telescope-
1. A large aperture is not required to focus incoming light, hence it is easy to manufacture.
2. No spherical aberration exists, and thus a clear image is formed.
An optical device used by watch repairers is __________.
Give reasons:A simple microscope is used by watch repairers.
A pencil partly immersed in water in a glass tumbler appears to be bent at the interface of air and water. Name the phenomenon of light responsible for it.
List three optical instruments, and describe what they do ?
Does the source appear to be nearer or farther with respect to surface AB ?
The source will appear nearer to the surface AB.
We can see from the figure that with increase in the thickness of the glass the shift $$OO'$$ increases.
Three convex lenses are available having focal lengths of 4 cm, 40 cm and 4 cm respectively. Which one would you choose as a magnifying glass and why?
A ray of light falls on three glass slabs of the same material at the same angle of incidence. In which case, is the lateral displacement of the emergent ray of light
(a) maximum
(b) minimum
Draw a ray diagram to illustrate the bending of a stick in water
Due to refraction from denser medium to rarer medium, the light coming from stick's end refracts away from normal as shown in figure
and thus when refracted ray extended backward then it meets sticks end at some height above to actual depth of end.
that's why the sticks appears bent and shorter inside water.
A coin is placed at the bottom of a tin can. Move your eyes until the coin is just out of sight (see diagram). Keep your eyes in this position while someone pours water gradually into the tin can. What would you observe ? Explain your observation.
When water is poured then it is due to refraction of light taking place that the coin becomes visible.
Let a coin be placed in an opaque container which is not visible since the light rays from the coin travelling in straight lines are blocked by the walls of the container. When water is poured, light travels in a straight line until it reaches the surface of water, then it bends. When produced backwards this ray appears to come from the position C '. Since the point C' is at a higher level than C, the coin appears raised and becomes visible.
With the help of a well-labelled diagram show that the apparent depth of an object, such as a coin, in water is less than its real depth.
Attached picture shows the difference between the real depth and apparent depth of the object under water.
We can see that objects closer than their real depth to the surface only if the rays coming from them reach our eyes. In this picture, ray coming from the coin reaches the observer’s eye after refraction and so observer sees the image of the coin at the distance D from the surface which is the apparent depth of the coin which is the result of the refraction of light.
What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass $$=$$ 1.Does the answer depend on the location of the slab?
A slab of refractive index $$1.5$$ is placed in water of refractive index $$\dfrac {4}{3}$$. If the apparent shift in the object $$O$$ due to the slab is $$1\ cm$$, the thickness of slab is
A Cassegrain telescope uses two mirrors as shown in above figure. Such a telescope is built with the mirrors $$20\,mm$$ apart. If the radius of curvature of the large mirror is $$220\,mm$$ and the small mirror is $$140\,mm$$, where will the final image of an object at infinity be?
A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b)
What
is the maximum and the minimum angular magnification (magnifying
power) possible using the above simple microscope?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
A card sheet divided into squares each of size $$1 \,mm^2$$ is being viewed at a distance of $$9 cm$$ through a magnifying glass (a converging lens of focal length $$9 cm$$) held close to the eye.(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?(b) What is the angular magnification (magnifying power) of the lens?(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain
At what angle should a ray of light be incident on the face of a prism of refracting angle $$60^\circ$$ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is $$1.524$$.
(i) A screen is placed at a distance of 100 cm from an object. The image of the object is formed on the screen by a convex lens for two different locations of the lens separated by 20 cm. Calculate the focal length of the lens used.
(ii) A converging lens is kept coaxially in contact with a diverging lens - both the lenses being of equal focal length. What is the focal length of the combination?
A magnifying glass is what kind of lens, and where do where should I place the lens relative to the object to see an image larger than the object and right-side-up? ("F" stands for the focal length of the lens.)
A convex lens of focal length 25 m is placed coaxially in contact with a concave lens of focal length 20 cm .Determine the power of the combination. Will the system be converging or diverging in nature?
Draw a labeled ray diagram of a refracting telescope. Define its magnifying power and write the expression for it.
Write two important limitation of a refracting telescope over a reflecting type telescope
Refracting telescope:
Magnifying power- The magnifying power is the ratio of the angle $$\alpha$$ subtended at the eye by the final image to the angle $$\beta$$ which the object subtends at the lens or the eye.
$$\displaystyle m=\frac{\beta}{\alpha}=\frac{h}{f_{e}}\frac{f_{0}}{h}=\frac{f_{0}}{f_{e}}$$
Limitations of refracting telescope over reflecting type telescope-
(i) Refracting telescope suffers from chromatic aberrations as it uses large sized lenses.
(iii) The requirement of big lenses tend to be very heavy and therefore difficult to make and support by their edges
Magnifying power- The magnifying power is the ratio of the angle $$\alpha$$ subtended at the eye by the final image to the angle $$\beta$$ which the object subtends at the lens or the eye.
$$\displaystyle m=\frac{\beta}{\alpha}=\frac{h}{f_{e}}\frac{f_{0}}{h}=\frac{f_{0}}{f_{e}}$$
Limitations of refracting telescope over reflecting type telescope-
(i) Refracting telescope suffers from chromatic aberrations as it uses large sized lenses.
(iii) The requirement of big lenses tend to be very heavy and therefore difficult to make and support by their edges
A convex lens of focal length $$f_{1}$$ is kept in contact with a concave lens of focal length $$f_{2}$$. Find the focal length of the combination.
Velocity of light in glass is $$2 \times 10^8 ms^{-1}$$ and that in air is $$3 \times 10^8 ms^{-1}$$. By how much would an ink dot appear to be raised, when covered by a glass plate 6.0 cm thick?
Take a glass vessel and pour some glycerine into it and then pour water up to the brim. Take a quartz glass rod. Keep it in the vessel. Observe the glass rod from the sides of the glass vessel.
What changes do you notice?
When this arrangement is viewed from side, the size of the glass rod appears increased in water and glass rod disappeared in glycerine.
Take a glass vessel and pour some glycerine into it and then pour water up to the brim. Take a quartz glass rod. Keep it in the vessel. Observe the glass rod from the sides of the glass vessel.
What could be the reasons for these changes?
When this arrangement is viewed from side, the size of the glass rod appears increased in water($$\mu=1.33$$) and glass rod($$\mu=1.5$$) disappeared in glycerine($$\mu=1.47$$) because both the glass rod and glycerine have same refractive index. When mediums have same refractive index the speed of light is same in both the medium, so no bending or refraction takes place.
The size of glass rod appears increased in water due to taking place of refraction.
At critical angle, the angle of refraction is _______.
Suppose you are inside the water in a swimming pool near an edge. A friend is standing on the edge. Do you find your friend taller or shorter than his usual height? Why?
Define critical angle. Explain total internal reflection using a neat diagram.
Draw a neat (labelled) diagram for the formation of image in a simple microscope.
Light from a light source (mirror) passes through a thin transparent object. A biconvex lens magnifies the size of the object to get an enlarged virtual image. The image is viewed from the other side.
What is chromatic aberration ? How can it be minimized or eliminated ?
Drive an expression for equivalent focal length of two thin lenses kept in contact.
Let the focal lengths of the two lenses be $$f_1$$ and $$f_2$$.
$$u_1 = OC_1$$ and
$$v_1 = I_1C_1$$
$$u_2= -C_1I_1$$ which is approximately equal to $$C_2 I_1$$ and
$$v_2 = C_2I$$ which is approximately equal to $$C_1I$$
$$\therefore \frac{1}{u_1} + \frac{1}{v_1} = \frac{1}{f_1}$$
and $$ \frac{1}{u_2} + \frac{1}{v_2} = \frac{1}{f_2}$$
$$\frac{1}{OC_1} + \frac{1}{I_1C_1} = \frac{1}{f1}$$ and
$$\frac {1}{- C_1I_1} + \frac{1}{C_1I} = \frac{1}{f_2}$$
So $$\frac{1}{OC_1} + \frac{1}{I_1C_1} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{F}$$.
Combined focal length of two thin lenses in contact is given by:
$$F=\frac{f_1f_2}{f_1+f_2}$$
$$u_1 = OC_1$$ and
$$v_1 = I_1C_1$$
$$u_2= -C_1I_1$$ which is approximately equal to $$C_2 I_1$$ and
$$v_2 = C_2I$$ which is approximately equal to $$C_1I$$
$$\therefore \frac{1}{u_1} + \frac{1}{v_1} = \frac{1}{f_1}$$
and $$ \frac{1}{u_2} + \frac{1}{v_2} = \frac{1}{f_2}$$
$$\frac{1}{OC_1} + \frac{1}{I_1C_1} = \frac{1}{f1}$$ and
$$\frac {1}{- C_1I_1} + \frac{1}{C_1I} = \frac{1}{f_2}$$
So $$\frac{1}{OC_1} + \frac{1}{I_1C_1} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{F}$$.
Combined focal length of two thin lenses in contact is given by:
$$F=\frac{f_1f_2}{f_1+f_2}$$
The focal length of objective lens and eyepiece for a compound microscope are $$0.95$$ and $$5$$ cm and they are placed at $$20$$ cm from each other final image is formed at $$25$$ cm in eyepiece find the magnifying power of the microscope?
Explain refraction of light and lateral shift using a rectangular glass slab with figure.
Consider a rectangular glass slab PQRS having parallel faces PQ and RS as shown in the figure. A ray of light AB in the air is incident on the glass surface PQ. As the ray AB enters from the air(rarer medium) to glass (denser medium), the ray bends towards the normal and follows the path BC inside the glass slab. At point C, refraction takes place again. As the ray BC goes out from glass(denser medium) to air (rarer medium), the ray bends away from the normal and follows the path CD outside the glass slab. Here, the ray AB is called the incident ray, BC the refracted ray and CD the emergent ray.
Let $$n_1$$ and $$n_3$$ be the refractive index of air and $$n_2$$ the refractive index of glass.
On applying smells law at the surface PQ,
$$n_1 sin\theta_1 = n_2 sin \theta_2$$
For air medium, $$n_1=1$$
$$\therefore sin \theta_1 = n_2 sin \theta_2$$ ......(1)
On applying snells law at the surface RS,
$$n_2 sin \theta_3=n_3 sin \theta_4$$
For air medium $$n_3=1$$
$$n_2 sin \theta_3=sin \theta_4$$ ......(2)
As $$\theta_3$$ and $$\theta_2$$ are alternate angles, we have $$\theta_3 =\theta_2$$
Therefore (2) becomes:
$$n_2 sin \theta_2=sin \theta_4$$ .......(3)
Comparing (1) and (3) we get:
$$sin\theta_1 =sin \theta_4$$
$$\therefore \theta_1=\theta_4$$
i.e., angle of incidence $$=$$ angle of emergence
The emergency ray CD is parallel to the incident ray ABEF. However, the emergent ray is linearly displaced by a perpendicular distance CE because of refraction. When a ray of light is refracted from two refracting surfaces, the emergent ray is displaced from the direction of the incident ray. This displacement of the emergent ray from the path of the incident ray is called lateral shift.
Lateral shift depends on
i. The perpendicular distance between the two parallel refracting surfaces.
ii. The angle of incidence.
iii. Refractive index of the medium.
Let $$n_1$$ and $$n_3$$ be the refractive index of air and $$n_2$$ the refractive index of glass.
On applying smells law at the surface PQ,
$$n_1 sin\theta_1 = n_2 sin \theta_2$$
For air medium, $$n_1=1$$
$$\therefore sin \theta_1 = n_2 sin \theta_2$$ ......(1)
On applying snells law at the surface RS,
$$n_2 sin \theta_3=n_3 sin \theta_4$$
For air medium $$n_3=1$$
$$n_2 sin \theta_3=sin \theta_4$$ ......(2)
As $$\theta_3$$ and $$\theta_2$$ are alternate angles, we have $$\theta_3 =\theta_2$$
Therefore (2) becomes:
$$n_2 sin \theta_2=sin \theta_4$$ .......(3)
Comparing (1) and (3) we get:
$$sin\theta_1 =sin \theta_4$$
$$\therefore \theta_1=\theta_4$$
i.e., angle of incidence $$=$$ angle of emergence
The emergency ray CD is parallel to the incident ray ABEF. However, the emergent ray is linearly displaced by a perpendicular distance CE because of refraction. When a ray of light is refracted from two refracting surfaces, the emergent ray is displaced from the direction of the incident ray. This displacement of the emergent ray from the path of the incident ray is called lateral shift.
Lateral shift depends on
i. The perpendicular distance between the two parallel refracting surfaces.
ii. The angle of incidence.
iii. Refractive index of the medium.
Draw a ray diagram showing the formation of image by a reflecting telescope.
A converging lens of focal length $$40$$cm is kept in contact with a diverging lens of focal length $$30$$cm. Find the focal length of the combination.
How is the refractive index of a material related to real and apparent depth?
The image of an object form at the distance of distinct vision from the lens of a simple microscope of focal length $$2.5cm$$. It magnifying power is ____.
An object is placed $$20\ cm$$ in front of a block of glass $$10\ cm$$ thick and its farther surface is silvered. The image is formed $$23.2\ cm$$ behind the silvered face. Determine the refractive index of the glass.
Three lenses of power $$+1.5\ D$$ and $$-0.5\ D$$ are kept in contact on their principal axis. What is the effective power of the combination?
Why must both the $$f_0$$ and the $$f_e$$ of a compound microscope have short focal length?
A vessel contains water up to a height of $$20\ cm$$ and above it an oil up to another $$20\ cm$$. The refractive indices of the water and the oil are $$1.33$$ and $$1.30$$ respectively. Find the apparent depth of the vessel when viewed from above.
A ray of light of wavelength 5400 A suffers refraction from air to glass. Taking $$a_{\mu}g$$ as 3/2, find the wavelength of light in glass.
Derive $$\frac { 1 }{ f } =\frac { 1 }{ v } +\frac { 1 }{ u } $$ formula for spherical mirror.
A narrow beam of light is incident at $$53^{\circ}$$ angle made with normal on a glass plate of refractive index $$1.6$$. If the thickness of the plate is $$20\;mm$$. Calculate the shift of the beam.
A convex lens of focal length $$f_1$$ is kept in contact with a concave lens of focal length $$f_2$$. Find the focal length of the combination.
A postage stamp kept below a rectangular glass slab of refractive index $$1.5$$ when viewed vertically above it, appears to be raised by $$7.0 mm$$. Calculate the thickness of the glass slab.
An object is placed in front of a glass slab $$(\mu = 1.5)$$ of thickness $$6 \,cm$$ at a distance $$28 \,cm$$ from it. Other face of glass slab is silvered. Find the position of final image of object.
A simple microscope is rate 5 X for a normal relaxed eye. What will be its magnifying power for a relaxed farsighted eye whose near point is 40?
A candle flame 3 cm high is placed at a distance of 3 m from a wall.how far from the wall must a concave mirror be placed so that it may form 9 cm high image of the flame on the same wall? Also find the focal length of the mirror.
What is the lateral displacement when a ray of light falls normally on a glass slab?
Does the magnifying power of a microscope depend on the colour of the light used? Justify your answer.
Define critical angle. Prove that the refractive index of the denser medium is reciprocal of the sine of the critical angle .
The critical angle is the angle of incidence where the angle of refraction $${90^0}.$$ The light must travel from an optically more dense medium to an optically less dense medium$$.$$
Prof$$:-$$
Instead of always having to measure the critical angles of different materials$$,$$ it is possible to calculate the critical angle at the surface between two media using snell's Law$$.$$ To recap$$,$$ sneel's Law states$$:$$
$${n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}$$
Where $${n_1}$$ is the refractive index of material $$1,$$ $${n_2}$$ is the refractive index of material $$2,$$ $${\theta _1}$$ is the angle of incidence and $${\theta _2}$$ is the angle of refraction $$.$$ For total internal reflection we know that the angle of incidence is the critical angle$$,$$ so$$,$$
$${\theta _1} = {\theta _c}.$$
However$$,$$ we also know that the angle of refraction at the critical angle is $${90^0}.$$ So we have
$${\theta _2} = {90^0}$$
WE can then write snell's Law as$$:$$
$${n_1}\sin {\theta _c} = {n_2}\sin {90^0}$$
Solving for $${\theta _c}$$ gives$$:$$
$${n_1}\sin {\theta _c} = {n_2}\sin {90^0}$$
$$\sin {\theta _c} = \frac{{{n_2}}}{{{n_1}}}\left( 1 \right)$$
$${\theta _c} = {\sin ^{ - 1}}\left( {\frac{{{n_2}}}{{{n_1}}}} \right)$$
Hence$$,$$ we proved$$.$$
What is this distance between the lens and the screen called ?
A particle executes a simple harmonic motion of amplitude $$1.0 cm$$ along the principal axis of convex lens of focal length $$12 cm$$. The mean position of oscillation is at $$20 cm$$ from the lens. Find the amplitude of oscillation of the image of the particle.
When the object is at $$19cm$$ from the lens, let the image will be at, $$v_1$$.
$$\Rightarrow \dfrac{1}{v_1} - \dfrac{1}{\mu_1} = \dfrac{1}{f}$$
$$\Rightarrow \dfrac{1}{v_1} - \dfrac{1}{-19} = \dfrac{1}{12}$$
$$\Rightarrow v_1 = 32.57cm$$
Again, when the object is at $$21cm$$ from the lens, let the image will be at $$v_2$$
$$\Rightarrow \dfrac{1}{v_2} - \dfrac{1}{u_2} = \dfrac{1}{f} $$
$$\Rightarrow \dfrac{1}{v_2} + \dfrac{1}{21} = \dfrac{1}{12}$$
$$\Rightarrow v_2 = 28cm$$
$$\therefore$$ Amplitude of vibration of the image is $$A = \dfrac{A'B'}{2} = \dfrac{v_1 - v_2}{2}$$
$$\Rightarrow A = \dfrac{32.57 - 28}{2} = 2.285cm$$
If the glass slab is of the following shape , and the experiment is performed as above , will the effects be same ? Here $$S_{ 1 }$$ and $$S_{ 2 }$$ are slab surfaces parallel to each other.
Define critical angle and polarising angle what is the relation between the two angles ?
An object of height 4 cm is placed at a distance of 15 cm infront of a concave lens of power -10D. Find the size of the image.
When an object is placed at a distance of 15 cm from a concave mirror its image is formed at 10 cm in front of the mirror. Calculate the focal length of the mirror.
A mountaineer on a peak observes a rainbow in a valley, caused by water droplets in the sir $$ 8\ km $$ away. The valley is $$2\ km$$ below the mountain peak and is entirely flat. What fraction of the complete circular arc of the rainbow is visible to him?
The light that represents the rainbow, reaches the mountaineer at an angle of $$40^{\circ}$$ (for violet) and $$42^{\circ}$$ (for red) with the eye-level. Since the outer edge is Red and is larger, lesser percentage of it will be visible as compared to Violet. So, we consider the outer edge of the Rainbow.
Let the Radius of the outer circle be $$R$$.
$$R = 8 \times \sin(42^{\circ})\ km = 5.35\ km$$
Now,
To measure the angle of the part that is not visble, let the angle between the vertical axis and the bottom point of the rainbow where it touches the ground be $$\phi$$.
Then, $$\cos(\phi)=\dfrac{2\ km}{R} = \dfrac{2\ km}{5.35\ km} = 0.374$$
$$\Rightarrow \phi = \cos^{-1}(0.374)$$
$$\therefore \phi \approx 68^{\circ}$$
Now, the fraction of visible part of rainbow = $$\dfrac{360^{\circ} - 2 \times 68^{\circ}}{360^{\circ}} = \dfrac{224^{\circ}}{360^{\circ}} = 62.2\%$$
Thus, $$62.2\%$$ of the rainbow will be visible to the mountaineer.
A material having an index of refraction $$ n $$ is surrounded by a vacuum and is in the shape of a quarter circle of radius $$ R $$. Alight parallel to the base of the material is incident from the left at a distance of $$ L $$ above the base and emerges out of the material at the angle $$ \theta $$. Determine an expression for $$ \theta $$.
Given:
Refractive index of material = $$n$$
Radius of the material = $$R$$
Height of Light ray from base = $$L$$
Now, for the first refraction:
$$\dfrac{\sin(r_1)}{\sin(i)} = \dfrac{n_1}{n_2}$$
$$\Rightarrow \sin(r_1) = \dfrac{n_1}{n_2}\sin(i)$$
But, $$n_1 = 1$$, $$n_2 = n$$, and $$\sin(i) = \dfrac{L}{R}$$ (from the triangle)
$$\Rightarrow \sin(r_1) = \dfrac{L}{nR}$$
Now, $$i = r_1 + r_2$$ (since the sum of opposite interior angles equals exterior angle)
$$\Rightarrow r_2 = i - r_1 $$
$$\Rightarrow \sin(r_2) = \sin(i - r_1) = \sin(i)\cos(r_1) - \cos(i)\sin(r_1)$$
Again, for the second refraction:
$$\dfrac{\sin(\theta)}{\sin(r_2)} = \dfrac{n_1}{n_2}$$
$$\Rightarrow \sin(\theta) = \dfrac{n_1}{n_2}\sin(r_2)$$
But, $$n_1 = n$$, $$n_2 = 1$$
$$\Rightarrow \sin(\theta) = n\sin(i - r_1) = n\left(\sin(i)\cos(r_1)-\cos(i)\sin(r_1)\right)$$
$$\Rightarrow \sin(\theta) = n\left(\dfrac{L}{R}\sqrt{1 - \left(\dfrac{L}{nR}\right)^2} - \dfrac{L}{nR}\sqrt{1 - \left(\dfrac{L}{R}\right)^2}\right)$$
$$\Rightarrow \sin(\theta) = \dfrac{L}{R^2}\sqrt{n^2R^2 - L^2} - \dfrac{L}{R^2}\sqrt{R^2 - L^2}\\[4pt]$$
$$\therefore \theta = \sin^{-1}\left[\dfrac{L}{R^2}\left(\sqrt{n^2R^2 - L^2} - \sqrt{R^2 - L^2}\right)\right]$$
An object is placed 15cm from (a) a converging mirror, and (b) a diverging mirror, of radius of curvature 20cm. (c) Calculate the image position and magnification in case.
(a) fig. formation of image by the concave mirror when the object is placed between its pole and focus.
(b) f=-20cm, u=-10cm, v=?
We know that
$$\displaystyle \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$
$$\displaystyle \Rightarrow \frac{1}{v}+\frac{1}{\left (-10 \right )}=\frac{1}{-20}$$
$$\displaystyle \Rightarrow \frac{1}{v}=-\frac{1}{20}+\frac{1}{10}=\frac{1}{20}$$
$$\therefore v=20 cm $$
(c) Characteristics of the image formed
(i) Image is virtual.
(ii) Image is erect.
Two narrow parallel slits separated by $$0.850\ mm$$ are illuminated by $$600\ nm$$ light, and the viewing screen is $$2.80\ m$$ away from the slits. (a) What is the phase difference between the two interfering waves on a screen at a point $$2.50\ mm$$ from the central bright fringe? (b) What is the ratio of the intensity at this point to the intensity at the centre of a bright fringe?
The main focal, length of the objective of a microscope is $$ f_{obj} = 3 \,mm $$ and of the eyepiece $$ f_{eye} = 5 \,cm. $$ An object is at a distance of $$ a= 3.1 \, mm $$ from the objective. Find the magnification of the microscope for a normal eye.
A simple microscope has a magnifying power of $$3.0$$ when the image is formed at the near point $$(25 cm)$$ of a normal eye. What will be its magnifying power if the image is formed at infinity?
A light ray enters a rectangular glass slab at angle $$ \theta_1 = 45^{\circ} $$ and emerges at an angle $$ \theta_2 = 76^{\circ} $$, as shown in the figure. (a) Determine the refractive index of refraction for the slab. (b) If the light ray enters the plastic at a point $$ L = 50\, cm $$ from the bottom edge, how long does it take the light ray to travel through the slab?
The objective is taken out of a telescope adjusted to infinity and replaced by a diaphragm with a diameter $$ D.$$ A screen shows a real image of the diaphragm having a diameter $$ d $$ at a certain distance from the eyepiece. What was the magnification of the telescope?
Where should an object be placed in order to use a convex lens as a magnifying glass ?
Give three examples of material that refract light rays . What happens to the speed of light rays when they enter these materials ?
What name is given to the ratio of sine of angle of incidence to the sine of angle of refraction ?
(a) With the help of a diagram , show that how when light falls obliquely on the side of a rectangular glass slab, the emergent ray is parallel to the incident ray .
(b) Show the lateral displacement of the ray on the diagram .
(c) State two factors on which the lateral displacement of the emergent ray depends.
(b) The lateral displacement is shown in the above diagram
(c) Factors on which the lateral displacement depends are:
(i) Angle of incidence
(ii) Thickness of glass slab
(iii) Refractive index of glass slab.
(c) Factors on which the lateral displacement depends are:
(i) Angle of incidence
(ii) Thickness of glass slab
(iii) Refractive index of glass slab.
Fill in the following blanks with suitable words :
(a) Parallel rays of light are refracted by a convex lens to a point called the ______
(b) The image in a convex lens depends upon the distance of the ____ from the lens.
Where is the object placed in reference to the principal focus of a magnifying glass, so as to see its enlarged image? Where is the image obtained?
You look down at a coin that
lies at the bottom of a pool of liquid
of depth $$d$$ and index of refraction n. Because you view with
two eyes, which intercept different
rays of light from the coin, you perceive the coin to be where
extensions of the intercepted rays
cross, at depth $$d_a$$ instead of $$d$$.
Assuming that the intercepted rays
in Figure are close to a vertical axis through the coin, show that $$d_a = d/n$$. (Hint: Use the small-angle
approximation $$\sin \theta \approx \tan \theta \approx \theta$$.)
In above figure, a light ray in air is incident at angle $$\theta_{1}$$ on a block of transparent plastic with an index of refraction of $$1.56$$. The dimensions indicated are $$H=2.00cm$$ and $$W=3.00cm$$. The light passes through the block to one of its sides and there undergoes reflection (inside the block) and possibly refraction (out into the air). This is the point of first reflection. The reflected light then passes through the block to another of its sidesa point of second reflection. If $$\theta_{1}=40^{0}$$, on which side is the point of (a) first reflection and (b) second reflection? If there is refraction at the point of (c) first reflection and (d) second reflection, give the angle of refraction; if not, answer none. If $$\theta_{1}=70^{0}$$, on which side is the point of (e) first reflection and (f) second reflection? If there is refraction at the point of (g) first reflection and (h) second reflection, give the angle of refraction; if not, answer none.
(a) We note that the upper-right corner is at an angle (measured from the point where the light enters, and measured relative to a normal axis established at that point the normal at that point would be horizontal in figure given in question) is at $$\tan ^{-1}\left ( 2/3 \right )=33.7^{0}$$. The angle of refraction is given by
$$n_{air}\sin 40^{0}=1.56\sin \theta _{2}$$.
which yields $$\theta _{2}=24.33^{0}$$ if we use the common approximation $$n_{air}=1.0$$, and yields $$\theta _{2}=24.34^{0}$$ if we use the more accurate value for $$n_{air}$$ found in Table below. The value is less than $$33.7^{0}$$, which means that the light goes to side 3.
(b) The ray strikes a point on side 3, which is $$0.643cm$$ below that upper-right corner, and then (using the fact that the angle is symmetrical upon reflection) strikes the top surface (side 2) at a point $$1.42cm$$ to the left of that corner. Since $$1.42cm$$ is certainly less than $$3cm$$ we have a self-consistency check to the effect that the ray does indeed strike side 2 as its second reflection (if we had gotten $$3.42 cm$$ instead of $$1.42 cm$$, then the situation would be quite different).
(c) The normal axes for sides 1 and 3 are both horizontal, so the angle of incidence (in the plastic) at side 3 is the same as the angle of refraction was at side 1. Thus,
$$1.56\sin 24.3^{0}=n_{air}\sin \theta_{air}\Rightarrow \theta _{air}=40^{0}$$.
(d) It strikes the top surface (side 2) at an angle (measured from the normal axis there, which in this case would be a vertical axis) of $$90^{0}-\theta _{2}=66^{0}$$, which is much greater than the critical angle for total internal reflection ($$\sin^{-1}\left ( n_{air}/1.56 \right )=39.9^{0}$$). Therefore, no refraction occurs when the light strikes side 2.
(e) In this case, we have $$n_{air}\sin 70^{0}=1.56\sin \theta _{2}$$,
which yields $$\theta _{2}=37.04^{0}$$ if we use the common approximation $$n _{air}=1.0$$, and yields $$\theta _{2}=37.05^{0}$$ if we use the more accurate value for $$n _{air}$$ found in Table below. This is greater than the $$33.7^{0}$$ mentioned above (regarding the upper-right corner), so the ray strikes side 2 instead of side 3.
(f) After bouncing from side 2 (at a point fairly close to that corner) it goes to side 3.
(g) When it bounced from side 2, its angle of incidence (because the normal axis for side 2 is orthogonal to that for side 1) is $$90^{0}-\theta _{2}=53^{0}$$, which is much greater than the critical angle for total internal reflection (which, again, is $$\sin^{-1}\left ( n_{air}/1.56 \right )=39.9^{0}$$. Therefore, no refraction occurs when the light strikes side 2.
(h) For the same reasons implicit in the calculation of part (c), the refracted ray emerges from side 3 with the same angle ($$70^{0}$$) that it entered side 1. We see that the occurrence of an intermediate reflection (from side 2) does not alter this overall fact: light comes into the block at the same angle that it emerges with from the opposite parallel side.
$$n_{air}\sin 40^{0}=1.56\sin \theta _{2}$$.
which yields $$\theta _{2}=24.33^{0}$$ if we use the common approximation $$n_{air}=1.0$$, and yields $$\theta _{2}=24.34^{0}$$ if we use the more accurate value for $$n_{air}$$ found in Table below. The value is less than $$33.7^{0}$$, which means that the light goes to side 3.
(b) The ray strikes a point on side 3, which is $$0.643cm$$ below that upper-right corner, and then (using the fact that the angle is symmetrical upon reflection) strikes the top surface (side 2) at a point $$1.42cm$$ to the left of that corner. Since $$1.42cm$$ is certainly less than $$3cm$$ we have a self-consistency check to the effect that the ray does indeed strike side 2 as its second reflection (if we had gotten $$3.42 cm$$ instead of $$1.42 cm$$, then the situation would be quite different).
(c) The normal axes for sides 1 and 3 are both horizontal, so the angle of incidence (in the plastic) at side 3 is the same as the angle of refraction was at side 1. Thus,
$$1.56\sin 24.3^{0}=n_{air}\sin \theta_{air}\Rightarrow \theta _{air}=40^{0}$$.
(d) It strikes the top surface (side 2) at an angle (measured from the normal axis there, which in this case would be a vertical axis) of $$90^{0}-\theta _{2}=66^{0}$$, which is much greater than the critical angle for total internal reflection ($$\sin^{-1}\left ( n_{air}/1.56 \right )=39.9^{0}$$). Therefore, no refraction occurs when the light strikes side 2.
(e) In this case, we have $$n_{air}\sin 70^{0}=1.56\sin \theta _{2}$$,
which yields $$\theta _{2}=37.04^{0}$$ if we use the common approximation $$n _{air}=1.0$$, and yields $$\theta _{2}=37.05^{0}$$ if we use the more accurate value for $$n _{air}$$ found in Table below. This is greater than the $$33.7^{0}$$ mentioned above (regarding the upper-right corner), so the ray strikes side 2 instead of side 3.
(f) After bouncing from side 2 (at a point fairly close to that corner) it goes to side 3.
(g) When it bounced from side 2, its angle of incidence (because the normal axis for side 2 is orthogonal to that for side 1) is $$90^{0}-\theta _{2}=53^{0}$$, which is much greater than the critical angle for total internal reflection (which, again, is $$\sin^{-1}\left ( n_{air}/1.56 \right )=39.9^{0}$$. Therefore, no refraction occurs when the light strikes side 2.
(h) For the same reasons implicit in the calculation of part (c), the refracted ray emerges from side 3 with the same angle ($$70^{0}$$) that it entered side 1. We see that the occurrence of an intermediate reflection (from side 2) does not alter this overall fact: light comes into the block at the same angle that it emerges with from the opposite parallel side.
Draw a ray diagram showing the path of rays of light when enters with oblique incidence (i) from air into water,. (ii) from water into air
Use Huygens' Principle to show how a plane wave front propagates from a denser to rarer medium. Hence, verify Snell's law of refraction.
We assume a plane wave front AB propagating in denser medium incident on the interface PP at angle i as shown in
Fig.
Fig.
Let T be the time taken by the wave front to travel a distance BC. If vi is the speed of the light in medium I.
So, $$BC = v_1 T$$
So, $$BC = v_1 T$$
In order to find the shape of the refracted wave front, we draw a sphere of radius $$AE = v_2 T$$ , where $$v_2$$ is the speed of light in medium II (rarer medium).
The tangent plane CE represents the refracted wave front
In $$\Delta ABC = sin \ i = \frac{BC}{AC} = \frac{v_1t}{AC}$$
The tangent plane CE represents the refracted wave front
In $$\Delta ABC = sin \ i = \frac{BC}{AC} = \frac{v_1t}{AC}$$
and in $$\Delta ACE = sin \ r = \frac{AE}{AC} = \frac{v_2t}{AC}$$
$$\therefore \ \ \frac{sin \ i}{sin \ r} \frac{BC}{AE} = \frac{v_1t}{v_2t} = \frac{v_1}{v_2}$$ ...(1)
$$\therefore \ \ \frac{sin \ i}{sin \ r} \frac{BC}{AE} = \frac{v_1t}{v_2t} = \frac{v_1}{v_2}$$ ...(1)
Let c be the speed of light in vacuum
So, $$\mu_1 = \frac{C}{v_1}$$ and $$ \mu_2 = \frac{C}{v_2}$$
$$\frac{\mu_2}{\mu_1} = \frac{v_1}{v_2}$$ ...(2)
So, $$\mu_1 = \frac{C}{v_1}$$ and $$ \mu_2 = \frac{C}{v_2}$$
$$\frac{\mu_2}{\mu_1} = \frac{v_1}{v_2}$$ ...(2)
From equations (1) and (2), we have
$$\frac{sin \ i}{sin \ r} = \frac{\mu_2}{\mu_1}$$
$$\mu_1 \ sin \ i = \mu_2 \ sin \ r$$
It is known as Snell's law.
$$\frac{sin \ i}{sin \ r} = \frac{\mu_2}{\mu_1}$$
$$\mu_1 \ sin \ i = \mu_2 \ sin \ r$$
It is known as Snell's law.
Fill in the blanks :
When a ray of light travels from water to air , it bends _______ the normal.
How is the refractive index of a medium related to the real and apparent depths of an object in that medium?
The ray diagrams given below show the paths of a ray of light travelling from a medium M into different media 1,2 and 3.
(a) In which of three media 1,2 or 3 in descending order of (i) faster, (ii) slower than in medium M
In the following figures, one lens is placed inside each box. State the nature of the lens. Complete the ray diagrams
Draw a neat labelled ray diagram to show the formation of image by a magnifying glass. State three characteristics of the image.
Let the object (AB) is situated between focal length and optical center of a convex lens then its image (A'B') will form on the same side of lens.
The image formed will be virtual, magnified and erect.
The image formed will be virtual, magnified and erect.
Where should be an object placed in front of convex lens so as to use it as a magnifier?
A student puts his pencil into an empty trough and observes the pencil from the position as indicated in the figure above.
(i) What change will be observed in the appearance of the pencil when water is poured into the trough?
(ii) Name the phenomenon which accounts for the above started observation.
(iii) Complete the diagram showing how the student's eye sees the pencil through
water.
(i) Part of the pencil which is immersed in water will look short and raised up.
(ii) The phenomena which is responsible for the above observation is refraction of light.
(iii) The required image is shown below:
Define lateral displacement.
Draw a ray diagram to show the appearance of a stick partially immersed in water. Explain your answer.
A stick partially immersed in water in a glass container appears bent or raised as shown in figure below. This happens because the rays appears to come from $$P’$$ (which is the virtual image of the tip $$P$$ of the stick) due to the refraction from denser medium (water) to rarer medium (air) at the surface separating two media.
Solve the following example.
An object of height 7 cm is kept at a distance of 25 cm in front of a concave mirror. The focal length of the mirror is 15 cm. At what distance from the mirror should a screen be kept so as to get a clear image? what will be the size and nature of the image?
Given object height $$(h_o) = 7 cm$$
Object distance $$(u) = -25 cm$$
Foeal length $$(f) = -15 cm$$
Using mirror formula,
$$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f} $$ ; $$v = $$ distance of image
$$\Rightarrow \dfrac{1}{v} = \dfrac{1}{25} - \dfrac{1}{15}$$
$$\therefore \dfrac{1}{v} = \dfrac{2.5}{75}$$
$$\Rightarrow v = -37.5 cm$$
The screen should be placed at $$37.5 cm$$ from the pole of the mirror and infront of the mirror.
Magnification $$(m) = \dfrac{h_i}{h_o} = -\dfrac{v}{4}$$
$$\Rightarrow \dfrac{h_i}{h_o} = -\left(\dfrac{-37.5}{-25} \right) = -1.5$$
$$\therefore h_i = -(1.5 \times 7) cm = -10.5 cm$$
So, size of image is $$10.5 cm$$ and it will be real, inverted and magnified.
Where is the telescope used? Discuss it with your friends and teachers.
Which lens has magnification less than $$1 ?$$
Now, look in the glass from the top. What can you see?
What type of lens is used in simple microscope ?
How does the pencil appear, when it is kept in glass full of water?
Light of wavelength 700 nm is incident on the face of a fused quartz prism $$ (n=1.458 $$ at 700 nm) at an incidence angle of $$ 75.0^{\circ}. $$ The apex angle of the prism is $$ 60.0^{\circ}. $$ Calculate the angle of incidence at the second surface.
The index of refraction at $$ 700 \mathrm{nm} $$ is $$ n(700 \mathrm{nm})=1.458 $$.
Refer to ANS. FIG. P35.29. Let
$$\theta_{3}+\beta=90.0^{\circ} \text { and } \theta_{2}+\alpha=90.0^{\circ}$$
then,$$\alpha+\beta+60.0^{\circ}=180^{\circ}$$
So
$$ \alpha+\beta+60.0^{\circ}=180^{\circ} $$
$$\Rightarrow \left(90.0^{\circ}-\theta_{2}\right)+\left(90.0^{\circ}-\theta_{3}\right)+60.0^{\circ}=180^{\circ} $$
$$\Rightarrow 60.0^{\circ}-\theta_{2}-\theta_{3}=0 $$
Refer to ANS. FIG. P35.29. Let
$$\theta_{3}+\beta=90.0^{\circ} \text { and } \theta_{2}+\alpha=90.0^{\circ}$$
then,$$\alpha+\beta+60.0^{\circ}=180^{\circ}$$
So
$$ \alpha+\beta+60.0^{\circ}=180^{\circ} $$
$$\Rightarrow \left(90.0^{\circ}-\theta_{2}\right)+\left(90.0^{\circ}-\theta_{3}\right)+60.0^{\circ}=180^{\circ} $$
$$\Rightarrow 60.0^{\circ}-\theta_{2}-\theta_{3}=0 $$
$$\Rightarrow 60.0^{\circ}-41.5^{\circ}=\theta_{3}=18.5^{\circ} $$
An object is placed in front of a concave mirror 20 cm away from it. If its focal length is 40 cm, locate the position of image and its nature.
An object of 5 cm is kept in front of a convex mirror having radius of curvature 30 cm at a distance of 10 cm. Calculate the position, size, and feature of object.
Light passes from air into flint glass at a nonzero angle of incidence. Is it possible for the component of its velocity perpendicular to the interface to remain constant? Explain your answer.
Yellow light of wavelength $$ 589 \mathrm{nm} $$ is used to view an object under a microscope. The objective lens diameter is $$ 9.00 \mathrm{mm} $$.
Suppose water fills the space between the object and the objective. What effect does this change have on the resolving power when $$ 589-\mathrm{nm} $$ light is used?
In given Figure shows a pencil partially immersed in a cup of water. Why does the pencil appear to be bent?
A ray of light, incident obliquely on a face of a rectangular glass slab placed in air, emerges from the opposite face parallel to the incident ray.
State two factors on which the lateral displacement of the emergent ray depends.
List the sign conventions that are followed in case of refraction of light through spherical lenses. Draw a diagram and apply these conventions in determining the nature and focal length of a spherical lens which forms three times magnified real image of an object placed $$16\ cm$$ from the lens.
Sign conventions for refraction of light through spherical lens are:
$$u = -16\ cm , m = -3 $$ (real image)
But $$ m = \dfrac{v}{u} $$
$$ = -3 $$
$$ \Rightarrow$$ $$ v = -3u$$
$$ = -3 \times (-16) $$
$$ = 48\ cm $$.
Using lens formula,
$$ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} $$
$$ = \dfrac{1}{48} - \dfrac{1}{-16} = \dfrac{1}{48} + \dfrac{1}{16} $$
$$ = \dfrac{1 + 3}{48} = \dfrac{4}{48} = \dfrac{1}{12} $$
$$ \Rightarrow$$ $$f = +12\ cm $$
So, focal length of the given spherical lens is $$12\ cm$$. The positive sign of focal length shows that the nature of spherical lens is convex.
- The object is always placed to the left of the lens so that incident light moves from left to right.
- All distances are to be measured from the optical centre of the lens.
- The distances measured in the direction of incident light (along +ve x-axis) will be taken as positive. while those measured to the left of the origin (along -ve x-axis) will be taken as negative.
- All measurements of heights above the principal axis (along +ve y-axis) will be considered as positive while below it (along -ve y-axis) will be taken as negative.
$$u = -16\ cm , m = -3 $$ (real image)
But $$ m = \dfrac{v}{u} $$
$$ = -3 $$
$$ \Rightarrow$$ $$ v = -3u$$
$$ = -3 \times (-16) $$
$$ = 48\ cm $$.
Using lens formula,
$$ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} $$
$$ = \dfrac{1}{48} - \dfrac{1}{-16} = \dfrac{1}{48} + \dfrac{1}{16} $$
$$ = \dfrac{1 + 3}{48} = \dfrac{4}{48} = \dfrac{1}{12} $$
$$ \Rightarrow$$ $$f = +12\ cm $$
So, focal length of the given spherical lens is $$12\ cm$$. The positive sign of focal length shows that the nature of spherical lens is convex.
Light in air (assume $$ n=1 $$ ) strikes the surface of a liquid of index of refraction $$ n_{\ell} $$ at the polarizing angle. The part of the beam refracted into the liquid strikes a submerged slab of material with refractive index $$ n $$ as shown in Figure $$ \mathrm{P} 38.$$ The light reflected from the upper surface of the slab is completely polarized. Find the angle $$ \theta $$ between the water surface and the surface of the slab as a function of $$ n $$ and $$ n_{\ell} $$
Refer to ANS. FIG. P38.65 above. Light strikes the liquid at the polarizing angle $$ \theta_{p} $$, enters the liquid at angle $$ \theta_{2} $$, and then strikes the slab at the angle $$ \theta_{3}, $$ which is equal to the polarizing angle $$ \theta_{p}^{\prime} . $$ The angle between the water surface and the surface of the slab, $$ \theta $$, is related to the other angles by (from the triangle)
$$\theta+\left(90^{\circ}+\theta_{2}\right)+\left(90^{\circ}-\theta_{3}\right)=180^{\circ} \quad \rightarrow \quad \theta=\theta_{3}-\theta_{2}$$
Also,
$$\begin{aligned}\theta_{p}+90^{\circ}+\theta_{2} &=180^{\circ} \\\theta_{p} &=90^{\circ}-\theta_{2}\end{aligned}$$
For the air-to-liquid interface,
$$ \mathrm{So} $$
$$\begin{aligned}\tan \theta_{p} &=\dfrac{n_{2}}{n_{1}}=\dfrac{n_{\text {liquid }}}{n_{\text {air }}}=\dfrac{n_{\ell}}{1}=\dfrac{\sin \theta_{p}}{\cos \theta_{p}}=\dfrac{\sin \left(90^{\circ}-\theta_{2}\right)}{\cos \left(90^{\circ}-\theta_{2}\right)} \\&=\dfrac{\cos \theta_{2}}{\sin \theta_{2}}=\dfrac{1}{\tan \theta_{2}} \\\tan \theta_{2} &=\dfrac{1}{n_{\ell}} \quad \rightarrow \quad \theta_{2}=\tan ^{-1}\left(\dfrac{1}{n_{t}}\right)\end{aligned}$$
For the water-to-slab interface,
$$\tan \theta_{3}=\tan \theta_{p}^{\prime}=\dfrac{n_{\mathrm{slab}}}{n_{\mathrm{liquid}}}=\dfrac{n}{n_{\ell}} \rightarrow \theta_{3}=\tan ^{-1}\left(\dfrac{n}{n_{\ell}}\right)$$
Therefore,$$\theta=\theta_{3}-\theta_{2} \rightarrow \theta=\tan ^{-1}\left(\dfrac{n}{n_{t}}\right)-\tan ^{-1}\left(\dfrac{1}{n_{t}}\right)$$
$$\theta+\left(90^{\circ}+\theta_{2}\right)+\left(90^{\circ}-\theta_{3}\right)=180^{\circ} \quad \rightarrow \quad \theta=\theta_{3}-\theta_{2}$$
Also,
$$\begin{aligned}\theta_{p}+90^{\circ}+\theta_{2} &=180^{\circ} \\\theta_{p} &=90^{\circ}-\theta_{2}\end{aligned}$$
For the air-to-liquid interface,
$$ \mathrm{So} $$
$$\begin{aligned}\tan \theta_{p} &=\dfrac{n_{2}}{n_{1}}=\dfrac{n_{\text {liquid }}}{n_{\text {air }}}=\dfrac{n_{\ell}}{1}=\dfrac{\sin \theta_{p}}{\cos \theta_{p}}=\dfrac{\sin \left(90^{\circ}-\theta_{2}\right)}{\cos \left(90^{\circ}-\theta_{2}\right)} \\&=\dfrac{\cos \theta_{2}}{\sin \theta_{2}}=\dfrac{1}{\tan \theta_{2}} \\\tan \theta_{2} &=\dfrac{1}{n_{\ell}} \quad \rightarrow \quad \theta_{2}=\tan ^{-1}\left(\dfrac{1}{n_{t}}\right)\end{aligned}$$
For the water-to-slab interface,
$$\tan \theta_{3}=\tan \theta_{p}^{\prime}=\dfrac{n_{\mathrm{slab}}}{n_{\mathrm{liquid}}}=\dfrac{n}{n_{\ell}} \rightarrow \theta_{3}=\tan ^{-1}\left(\dfrac{n}{n_{\ell}}\right)$$
Therefore,$$\theta=\theta_{3}-\theta_{2} \rightarrow \theta=\tan ^{-1}\left(\dfrac{n}{n_{t}}\right)-\tan ^{-1}\left(\dfrac{1}{n_{t}}\right)$$
The image of a candle flame placed at a distance of $$45\ cm$$ from a spherical lens is formed on a screen placed at a distance of $$90\ cm$$ from the lens. Identify the type of lens and calculate its focal length. If the height of the flame is $$2\ cm$$, find the height of its image.
Light is incident on a prism with an equilateral triangular cross-section, making an angle of $$\theta$$ w.r.t the normal as shown in the figure. The index of refraction of the prism is $$\displaystyle \frac{2}{\sqrt{3}}$$. If the angle $$\theta$$ starts at $$60^{\circ}$$ and is gradually made smaller until no light exist from side BC, the final value of $$\theta$$ in degrees is
$$A = 60^\circ$$
$$r' + r = A$$
So, $$\sin 90^\circ = \sqrt2 \sin r$$
$$\Rightarrow r' = 45^\circ$$
$$45 + r = A$$
$$\Rightarrow r = 15^\circ$$
Since
$$(1) \sin\theta = \sqrt2 \sin 15^\circ$$
$$\sin\theta = \sqrt2 \sin 15^\circ$$
$$\theta = \sin^{-1}(\sqrt2 \sin 15^\circ)$$
The distance of the object from the lens is 1/x. Find out the value of x?
For a prism of refracting angle A and refractive indexAssume rays are incident at all angles of incidence $$0^o \leq i \leq 90^o$$. Ignore partial reflection
Using $$\sin \theta_c = \dfrac{1}{n_2}$$
$$\implies \theta_c = 30^o$$ (As $$n_2= 2$$)
Here $$\theta_c = r_2$$
This suggests that $$-$$
1) if $$r_2 = 30^o, $$ then the ray will graze on the face LN of the prism.
2) if $$r_2 < 30^o$$, then the ray will always be refracted from the face LN.
3) if $$r_2 > 30^o$$, then always total internal reflection will occur.
3) if $$r_2 > 30^o$$, then always total internal reflection will occur.
Using snell' law for face LM, $$1 \times sin i= 2 sin r_1 $$
Thus for $$ 0^o \leq i \leq 90^o, \implies 0^o \leq r_1 \leq30^o $$
As $$A = r_1 + r_2 $$
Now for Case A; $$A = 15 ^o$$
$$ r_1 + r_2 = 15^o \implies 0 ^o\leq r_2 \leq 15^o $$
Hence all the rays will refract into the air.
For Case B; $$A = 45 ^o$$
$$ r_1 + r_2 = 45^o \implies 15 ^o\leq r_2 < 45^o $$
Hence some rays will refract and some will reflect.
For Case C; $$A = 70 ^o$$
$$ r_1 + r_2 = 70^o \implies 40 ^o\leq r_2 < 70^o $$
Hence all the rays will be reflected back from face LN.
For Case D; $$A = 50 ^o$$
$$ r_1 + r_2 = 50^o \implies 20 ^o\leq r_2 < 50^o $$
Hence some rays will refract and some will reflect.
As shown in the figure, light is incident normally on one face of the prism. A liquid of refractive index $$\mu$$ is placed on the horizontal face $$AC$$. The refractive index of prism is $$3/2$$. If total internal reflection taken place on face $$AC$$, $$\mu$$ should be less than $$\dfrac {I\sqrt {3}}{4}$$, where $$I$$ is an integer. Find the value of $$I$$.
Critical angle between glass and liquid.
$$\sin \theta_{C} = \dfrac {\mu}{(3/2)} = \dfrac {2\mu}{3}$$
Angle of incidence on $$AC = 60^{\circ}$$
For $$TIR,\ \ \ i > \theta_{C}$$
$$\sin \theta_{C} = \dfrac {\mu}{(3/2)} = \dfrac {2\mu}{3}$$
Angle of incidence on $$AC = 60^{\circ}$$
For $$TIR,\ \ \ i > \theta_{C}$$
$$ \Rightarrow \sin 60^{\circ} >\sin \theta_c=\dfrac {2}{3}\mu$$
$$\mu < \dfrac {3\sqrt {3}}{4} = \dfrac {I\sqrt {3}}{4}$$ (given)
So, $$I = 3$$.
The figure given below shows a solid glass block of refractive index $$1.5$$. The lower surface of the block is silvered and behaves as a reflecting mirror. A white light ray $$AB$$ incidents on the glass block at an angle of incidence $$40^{\circ}$$. Copy the diagram and trace the path of reflected and refracted rays from the upper and lower surface of the block.
From the above diagram we can find all the unknown
If reflection occurs at surface PS then
$$\angle i=\angle r\\ r=40^{ 0 }$$ reflection angle
By snell's law
$$\mu =\frac { Sini }{ Sinr } \\ 1.5=\frac { sin{ 40 }^{ 0 } }{ sinr } $$
$$sinr=0.428\\ \Rightarrow r={ 25.34 }^{ 0 }\simeq { 26 }^{ 0 }$$
second when refracted light from B goes to surface QR and reflects back and come back to surface PS here both refraction and reflection occurs.
thus,
$$\frac { 1 }{ \mu } =\frac { Sinr }{ Sine } \\ \frac { 1 }{ 1.5 } =\frac { sin{ 26 }^{ 0 } }{ sine } \\ sine=0.642\\ \Rightarrow e=40^{ 0 }$$
A ray of light strikes a glass slab 5 cm thick, making an angle of incidence equal to $$30^{\circ}$$. Construct the ray diagram showing showing the emergent ray and the refracted ray through the glass block. The refractive index of glass is 1.Also, measure the lateral displacement of the ray. Take $$\sin 19.5^{\circ}\, =\, 1/3$$.
Refractive index (n) = 1.5,
angle of incidence (i) - $$30^{\circ}$$
By Snell's law, $$\displaystyle \frac {sin\, i}{sin\, r}\, =\, n$$
$$\therefore\, sin\, r\, =\, \displaystyle \frac {sin\, i}{n}\, =\, \displaystyle \frac {sin\, 30^{\circ}}{1.5}\, =\, \displaystyle \frac {1/2}{1.5}\, =\, \displaystyle \frac {1}{3}$$
$$\Rightarrow\, \angle r\, =\, 19.5^{\circ}$$ (since sin $$19.5^{\circ}\, =\, \displaystyle \frac {1}{3}$$)
By measurement, on proper scale lateral displacement = 1 cm nearly.
angle of incidence (i) - $$30^{\circ}$$
By Snell's law, $$\displaystyle \frac {sin\, i}{sin\, r}\, =\, n$$
$$\therefore\, sin\, r\, =\, \displaystyle \frac {sin\, i}{n}\, =\, \displaystyle \frac {sin\, 30^{\circ}}{1.5}\, =\, \displaystyle \frac {1/2}{1.5}\, =\, \displaystyle \frac {1}{3}$$
$$\Rightarrow\, \angle r\, =\, 19.5^{\circ}$$ (since sin $$19.5^{\circ}\, =\, \displaystyle \frac {1}{3}$$)
By measurement, on proper scale lateral displacement = 1 cm nearly.
A card sheet divided into squares each of size $$1 \,mm^2$$ is being viewed at a distance of $$9\, cm$$ through a magnifying glass (a converging lens of focal length $$9\, cm$$) held close to the eye.In order to view the squares distinctly with the maximum possible magnifying power.(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?(b) What is the angular magnification (magnifying power) of the lens?(c) Is the magnification in (a) equal to the magnifying power in (b)?Explain:
A card sheet divided into squares each of size $$1 \,mm^2$$ is being viewed at a distance of $$9 cm$$ through a magnifying glass (a converging lens of focal length $$9 cm$$) held close to the eye and the magnifying glass if the virtual image of each square is to have an area of $$6.25 mm^{2}$$. Would you be able to see the squares distinctly with your eyes very close to the magnifier?
A ray of light passes through a right angled prism as shown in figure. State the angles of incidence at the faces $$AC$$ and $$BC$$.
To find
angle of incident foces $$AC$$ & $$BC$$
The light ray incident on $$AC$$
$$i=45^{o}$$
& Light fall normally on $$BC$$ (Perpendicularly)
i.e angle of incident is zero for $$BC$$
$$i=0$$
A ray of light strikes a glass slab of thickness $$t$$
(a) Prove that the ray is deflected laterally by distance
$$t \sin \theta \left[ 1-\sqrt {\dfrac {1-\sin^{2} \theta}{n^{2}-\sin^{2}\theta}}\right]$$
(b) Prove that for a small angle of incident the lateral shift $$y$$ is given by $$y=t \theta \left( 1-\dfrac {1}{n}\right)$$
When $$n$$ is the refractive index of glass with respect to air
(a) Lateral shift, $$y=BC=AB \sin r=AB \sin (\theta-r)$$
$$ \Rightarrow y=\dfrac { t }{ \cos { r } } \left( \sin { \theta } \cos { r } -\cos { \theta } \sin { r } \right) =t\left( \sin { \theta } -\cos { \theta } \tan { r } \right) $$
$$=t \sin \theta (1-\cot \theta \tan r)$$ $$(1)$$
Where $$AB=t/\cos r$$ and $$r=$$ angle of refraction given by using Snells Law,
$$\dfrac {\sin \theta}{\sin r}=n$$
$$\Rightarrow \sin r=\dfrac { \sin \theta }{ n } $$ & $$\cos { r } =\sqrt { \dfrac { { n }^{ 2 }-\sin ^{ 2 }{ \theta } }{ n } } $$
Putting the values of $$\sin r$$ & $$\cos r$$ in $$(1)$$ we obtain
$$y=t\sin \theta \left[ 1-\cot { \theta } \dfrac { \dfrac { \sin \theta }{ n } }{ \sqrt { \dfrac { { n }^{ 2 }-\sin ^{ 2 }{ \theta } }{ n } } } \right] =t\sin { \theta } \left[ 1-\dfrac { \cos { \theta } }{ \sqrt { { n }^{ 2 }-\sin ^{ 2 }{ \theta } } } \right] $$
$$=t\sin \theta \left[ 1-\sqrt { \dfrac { 1-\sin ^{ 2 }{ \theta } }{ { n }^{ 2 }-\sin ^{ 2 }{ \theta } } } \right] $$
(b)If $$\theta$$ is small, $$\sin^{2} \theta$$ can be ignored in comparison to $$1$$, $$n^{2}$$
$$\Rightarrow \quad y\cong t\theta \left[ 1-\sqrt { \dfrac { 1 }{ { n }^{ 2 } } } \right] $$
$$\Rightarrow \quad y=t\theta \left[ 1-\dfrac { 1 }{ { n } } \right] $$.
$$ \Rightarrow y=\dfrac { t }{ \cos { r } } \left( \sin { \theta } \cos { r } -\cos { \theta } \sin { r } \right) =t\left( \sin { \theta } -\cos { \theta } \tan { r } \right) $$
$$=t \sin \theta (1-\cot \theta \tan r)$$ $$(1)$$
Where $$AB=t/\cos r$$ and $$r=$$ angle of refraction given by using Snells Law,
$$\dfrac {\sin \theta}{\sin r}=n$$
$$\Rightarrow \sin r=\dfrac { \sin \theta }{ n } $$ & $$\cos { r } =\sqrt { \dfrac { { n }^{ 2 }-\sin ^{ 2 }{ \theta } }{ n } } $$
Putting the values of $$\sin r$$ & $$\cos r$$ in $$(1)$$ we obtain
$$y=t\sin \theta \left[ 1-\cot { \theta } \dfrac { \dfrac { \sin \theta }{ n } }{ \sqrt { \dfrac { { n }^{ 2 }-\sin ^{ 2 }{ \theta } }{ n } } } \right] =t\sin { \theta } \left[ 1-\dfrac { \cos { \theta } }{ \sqrt { { n }^{ 2 }-\sin ^{ 2 }{ \theta } } } \right] $$
$$=t\sin \theta \left[ 1-\sqrt { \dfrac { 1-\sin ^{ 2 }{ \theta } }{ { n }^{ 2 }-\sin ^{ 2 }{ \theta } } } \right] $$
(b)If $$\theta$$ is small, $$\sin^{2} \theta$$ can be ignored in comparison to $$1$$, $$n^{2}$$
$$\Rightarrow \quad y\cong t\theta \left[ 1-\sqrt { \dfrac { 1 }{ { n }^{ 2 } } } \right] $$
$$\Rightarrow \quad y=t\theta \left[ 1-\dfrac { 1 }{ { n } } \right] $$.
Consider the situation in figure. The bottom of the pot is a reflecting plane mirror, $$S$$ is a small fish, and $$T$$ is a human eye. Refractive index of water is $$\mu$$.
A) At what distance(s) from itself will the fish see the image(s) of the eye?
B) At what distance(s) from itself will the eye see the images(s) of the fish?
a) Let X= distance of the image of the eye formed above the surface as seen by the fish
So, $$\dfrac { H }{ x } =\dfrac { \text{Real depth} }{ \text{Apparent depth} } =\dfrac {1 }{ \mu } $$ or $$X=\mu H$$
So, distance of the direct image $$=\dfrac {H }{ 2 } +\mu H=H\left(\mu+\dfrac { 1 }{ 2 } \right)$$
Similarly, Image through mirror $$=\dfrac { H }{ 2 } +(H+x)=\dfrac { 3H }{ 2 } =\mu H=H\left(\mu+\dfrac { 3 }{ 2 } \right)$$
b) Here, $$\dfrac { H/2 }{ y } =\mu$$, so, $$y=\dfrac { H }{ 2\mu } $$
Where, y= distance of the image of fish below the surface as seen by eye.
So, Direct image $$=H+y=H+\dfrac { H }{ 2\mu } =H\left(1+\dfrac { 1 }{ 2\mu } \right)$$
Again another image of fish will be formed $$H/2$$ below mirror.
So, the real depth for that image of fish becomes $$H+H/2=3H/2$$
So, Apparent from the surface of water $$=3H/2\mu$$
So, distance of the image from the eye = $$H+\dfrac {3H }{ 2\mu}=H\left(1+\dfrac {3 }{ 2\mu }\right )$$
With the help of ray diagram, describe the construction, working of a compound microscope when the final image is formed at least distance of distinct vision $$(D = 25 \,cm)$$. Derive an expression fords magnifying power (m).
Construction of compound microscope
A compound microscope consists of two convex lenses: an objective lens O of small aperture and an eye piece E of large aperture.The lens which is placed towards the object is called objective lens, while the lens which is towards our eye is called eye piece. These two convex lenses i.e. the objective and the eye piece have short focal length and are fitted at the free ends of two sliding tubes at a suitable distance from each other. Although the focal length of both the objective lens and eye piece is short, but the focal length of the objective lens O is a little shorter than that of the eye piece E. The reason for using the eye piece of large focal length and large aperture in a compound microscope is, so that it may receive more light rays from the object to be magnified and form a bright image.
A compound microscope consists of two convex lenses: an objective lens O of small aperture and an eye piece E of large aperture.The lens which is placed towards the object is called objective lens, while the lens which is towards our eye is called eye piece. These two convex lenses i.e. the objective and the eye piece have short focal length and are fitted at the free ends of two sliding tubes at a suitable distance from each other. Although the focal length of both the objective lens and eye piece is short, but the focal length of the objective lens O is a little shorter than that of the eye piece E. The reason for using the eye piece of large focal length and large aperture in a compound microscope is, so that it may receive more light rays from the object to be magnified and form a bright image.
Working of compound microscope
The ray diagram to show the working of compound microscope is shown in figure. A tiny object AB to be magnified is placed in front of the objective lens just beyond its principal focus fo’. In this case, the objective lens O of the compound microscope forms a real, inverted and enlarged image A’B’ of the object. Now A’B’ acts as an object for the eye piece E, whose position is adjusted so that A’B’ lies between optical centre C2 and the focus fe’ of eye piece. Now the eye piece forms a final virtual, inverted and highly magnified image A”B”. this final image A”B” is seen by our eye hold close to eye piece, after adjusting the final image A”B” at the least distance of distinct vision of 25 cm from the eye.
A parallel beam of light enters a glass hemisphere perpendicular to the flat face, as shown in Fig. 1.69. The radius is $$ R = 6.00 \,cm $$ and the index of refraction is $$ n = 1.560 $$. Determine the point at which the beam is focussed. (Assume paraxial rays).
A biconvex lens $$ (f_1 = 10.0\,cm) $$ is placed $$ 40.0 \,cm $$ infront of a concave mirror $$ (f_2 = 7.50 \,cm) $$ as shown in Fig. 1.An object $$ 2.00 \,cm $$ high is placed $$ 20.0 \,cm $$ to the left of the lens. Using the lens and mirror equations and a ray diagram, find the locations of three images : (a) the image formed by the lens as rays travel to the right, (b) the image formed after the rays reflect from the mirror, and (c) the final image after the leftward travelling rays once again pass through the lens. Complete the ray diagram and characterize each image and object.
In the above figure, name the ray which represents the correct path of light while emerging out through a glass block.
The mixture a pure liquid and a solution in a long vertical column (ie., horizontal dimensions<< vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in traveling a horizontal distance $$d<< h$$, the height of the column
Write laws of refraction. Explain the same with the help of ray diagram, when a ray of light passes through a rectangular glass slab.
Laws of refraction are as follows:
(a) The incident ray, the refracted ray and normal at the point of incidence; all lie in the same plane.
(b) The ratio of sine of incidence to sine of refraction is a constant, for the light of a given colour and for a given pair of media. This law is also called Snell's Law.
$$\bullet $$ABCD is a glass slab. EF is incident ray which is incident on point 0 on air-glass interface.
$$\bullet $$NO is normal and $$\angle E O N=\angle t_{1} ;$$ which is angle of incidence.
$$\bullet $$ N'O' is normal extended towards the glass slab and $$\angle N^{\prime} O O=\angle r_{1} ;$$ which is angle of refraction.
$$\bullet $$ OO' is refracted ray from surface AB. It behaves like incident ray on surface CD.
$$\bullet $$ the ray EF bends when it enters the slab to become 00:
$$\bullet $$ MO' and O'M' are normal on surface CD.
$$\bullet $$GH is the emergent ray.$$\bullet $$ $$\angle O
O^{\prime} M=\angle i_{2}$$; which is angle of incidence at surface $$\mathrm{CD}$$
$$\bullet $$ $$\angle M
O^{\prime} H=\angle r_{2}$$; which is angle of refraction at surface $$C D$$.
$$\bullet $$ It ismobserved that EF, NO and 00' lie in the same plane; which is in accordance to the first law of refraction.
$$\bullet $$ It is also observed that $$E F \| G H$$; which means emergent ray is parallel to incident ray. This happens because the degree of bend at opposite surfaces of glass slab is same
A ray of light is incident normally on a plane glass slab. What will be (i) the angle of refraction and (ii) the angle of deviation for the ray?
What do you understand by refraction? Write the laws of refraction. Explain refraction through a glass slab.
When a ray of light travels from one medium to another medium there is a deviation in its path. This phenomenon is called refraction of light. Following are the laws of refraction:
- When a ray of light travels from a rarer medium to a denser medium, it bends towards the normal.
- When a ray of light travels from a denser medium to a rarer medium, it bends away from the normal.
- Ratio of sin of angle of incident to sin of angle reflection is constant for a given pair of media. This is called Snell's law.
- Fix a sheet of white paper on a drawing board using drawing pins.
- Place a rectangular glass slab over the sheet in the middle.
- Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
- Take four indentical pins.
- Fix two pins, E and F vertically such that the line joining the pins is inclined to the edge AB.
- Look for the images of the pins E and F through the opposite edge. Fix two pins, say G and H, such that these pins and the images of E and F lie on a straight line.
- Remove the pins and the slab.
- Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to edge CD. Let HG meet CD at 'O'.
- Join O and O'. Also produce EF up to P, as shown by a dotted line.
Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram.
When a ray of light enters from one rarer medium (say air) to another denser medium (say glass), it bends towards the normal. When the same ray of light exits from second medium into first medium, it bends away from the normal. In this case, the extent of bending of ray at opposite parallels (air-glass interface and glass-air interface) is same. Due to this, the emergent ray is parallel to incident ray.
(i) Drive the mirror formula. What is the corresponding formula for a thin lens?
(ii) Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the pole. Use this diagram, drive the magnification formula for the image formed.
What will be the angle of refraction for side $$RS$$?
With the necessary figure, explain the refraction of light through a rectangular glass slab.
Consider a rectangular glass slab ABCD having parallel faces AB and CD as shown in above figure. A ray of light EF in air is incident on the glass surface AB at point O. As the ray EO enters from air (rarer medium) to glass (denser medium), the ray bends towards normal and follows the path OO' inside the glass slab. At point O', refraction takes place again. As the ray OO' enters from glass (denser medium) to air (rarer medium), the ray bends away from normal and follow the path O'H outside glass slab.
Here, the ray EF is called incident ray, OO', the refracted ray and O'H, the emergent ray.
As per the laws of refraction and $$\angle i_1 = \angle r_1$$ and $$\angle i_2 = \angle r_2$$.
The emergent ray O'H is parallel to the incident ray EF. The perpendicular distance O'L between the original path of incident ray and the emergent ray is called the lateral displacement.
Here, the ray EF is called incident ray, OO', the refracted ray and O'H, the emergent ray.
As per the laws of refraction and $$\angle i_1 = \angle r_1$$ and $$\angle i_2 = \angle r_2$$.
The emergent ray O'H is parallel to the incident ray EF. The perpendicular distance O'L between the original path of incident ray and the emergent ray is called the lateral displacement.
What happens if light falls on a glass slab making $$ 90^{\circ} $$ at its surface?
Light in air strikes a water surface at the polarizing angle. The part of the beam refracted into the water strikes a submerged slab of material with refractive index $$ n=1.62 $$ as shown in Figure $$ \mathrm{P} 38.$$ The light reflected from the upper surface of the slab is completely polarized. Find the angle $$ \theta $$ between the water surface and the surface of the slab.
In ANS. FIG. P38.65, light strikes the liquid at the polarizing angle $$ \theta_{p \prime} $$ enters the liquid at angle $$ \theta_{2}, $$ and then strikes the slab at the angle $$ \theta_{3} $$, which is equal to the polarizing angle $$ \theta_{p}^{\prime} . $$ The angle between the water surface and the surface of the slab, $$ \theta $$, is related to the other angles by (from the triangle)
$$\theta+\left(90^{\circ}+\theta_{2}\right)+\left(90^{\circ}-\theta_{3}\right)=180^{\circ} \quad \rightarrow \quad \theta=\theta_{3}-\theta_{2}$$
For the air-to-water interface,
$$\tan \theta_{p}=\dfrac{n_{\text {water }}}{n_{\text {air }}}=\dfrac{1.33}{1.00} \rightarrow \theta_{p}=53.1^{\circ}$$
and
$$\begin{array}{l}(1.00) \sin \theta_{p}=(1.33) \sin \theta_{2} \\\theta_{2}=\sin ^{-1}\left(\dfrac{\sin 53.1^{\circ}}{1.33}\right)=36.9^{\circ}\end{array}$$
For the water-to-slab interface,
$$\tan \theta_{3}=\tan \theta_{p}=\dfrac{n_{\mathrm{s} 4 \mathrm{b}}}{n_{\mathrm{water}}}=\dfrac{n}{1.33}=\dfrac{1.62}{1.33}$$
$$\theta_{3}=50.6^{\circ}$$
The angle between surfaces is $$ \theta=\theta_{3}-\theta_{2}=13.7^{\circ} $$
$$\theta+\left(90^{\circ}+\theta_{2}\right)+\left(90^{\circ}-\theta_{3}\right)=180^{\circ} \quad \rightarrow \quad \theta=\theta_{3}-\theta_{2}$$
For the air-to-water interface,
$$\tan \theta_{p}=\dfrac{n_{\text {water }}}{n_{\text {air }}}=\dfrac{1.33}{1.00} \rightarrow \theta_{p}=53.1^{\circ}$$
and
$$\begin{array}{l}(1.00) \sin \theta_{p}=(1.33) \sin \theta_{2} \\\theta_{2}=\sin ^{-1}\left(\dfrac{\sin 53.1^{\circ}}{1.33}\right)=36.9^{\circ}\end{array}$$
For the water-to-slab interface,
$$\tan \theta_{3}=\tan \theta_{p}=\dfrac{n_{\mathrm{s} 4 \mathrm{b}}}{n_{\mathrm{water}}}=\dfrac{n}{1.33}=\dfrac{1.62}{1.33}$$
$$\theta_{3}=50.6^{\circ}$$
The angle between surfaces is $$ \theta=\theta_{3}-\theta_{2}=13.7^{\circ} $$
A beam of light both reflects and refracts at the surface between air and glass as shown in this Figure. If the refractive index of the glass is $$ n_{g}, $$ find the angle of incidence $$ \theta_{1} $$ in the air that would result in the reflected ray and the refracted ray being perpendicular to each other.
A light ray enters a rectangular block of plastic at an angle $$ \theta_{1}=45.0^{\circ} $$ and emerges at an angle $$ \theta_{2}=76.0^{\circ} $$ as shown in this Figure.
Determine the index of refraction of the plastic.
Given that $$ \theta_{1}=45.0^{\circ} $$ and $$ \theta_{2}=76.0^{\circ} $$,
Snell's law at the first surface gives
$$n \sin \alpha=1.00 \sin 45.0^{\circ} \quad[1]$$
Observe that the angle of incidence at the second surface is
$$\beta=90.0^{\circ}-\alpha$$
Thus, Snell's law at the second surface yields
$$\begin{aligned}& n \sin \beta=n \sin \left(90.0^{\circ}-\alpha\right)=1.00 \sin 76.0^{\circ} \\\text { or } & n \cos \alpha=\sin 76.0^{\circ} \quad[2]\end{aligned}$$
Dividing equation [1] by equation [2], we obtain
$$\tan \alpha=\dfrac{\sin 45.0^{\circ}}{\sin 76.0^{\circ}}=0.729$$
or $$ \alpha=36.1^{\circ} $$
Then, from equation [1]
$$n=\dfrac{\sin 45.0^{\circ}}{\sin \alpha}=\dfrac{\sin 45.0^{\circ}}{\sin 36.1^{\circ}}=1.20$$
Draw a neat diagram showing Refraction of light through a rectangular glass slab
A train fo plane light waves propagates in the medium where the phase velocity $$v$$ is a linear function of wavelength: $$v = a + b\lambda$$, where $$a$$ and $$b $$are some positive constants. Demonstrate that in such a medium the shape of an arbitrary train of light waves is restored after the time interval $$\tau = 1/b$$.
A light beam containing red and violet wavelengths is incident on a slab of quartz at an angle of incidence of $$ 50.0^{\circ}. $$ The index of refraction of quartz is 1.455 at $$ 600 \mathrm{nm} $$ (red light), and its index of refraction is 1.468 at $$ 410 \mathrm{nm} $$ (violet light). Find the dispersion of the slab, which is defined as the difference in the angles of refraction for the two wavelengths.
Light passes from air into flint glass at a nonzero angle of incidence. What If? Can the component of velocity parallel to the interface remain constant during refraction? Explain your answer.
A beam of natural light intensity $$I_{0}$$ falls on a system of two crossed Nicol prism between which a tube filled with a certain solution is placed in a longitudinal magnetic field of strength $$H$$. The length of the tube is $$l$$, the coefficient of linear absorption of solution is $$x$$, and the Verdict constant is $$V$$. Find the intensity of the light transmitted through that system.
(a)With the help of a ray diagram explain why a concave lens diverges the rays of a parallel beam of light.
(b) A $$2.0\ cm$$ tall object is placed perpendicular to the principal axis of a concave lens of focal length $$15\ cm$$. At what distance from the lens, should the object be placed so that it forms an image $$10\ cm$$ from the lens? Also find the nature and the size of image formed.
(a) When a parallel beam of light incident on a front face of concave lens, each ray of light will refract towards the normal to the surface as it moves from rarer to denser medium and travels in a straight line inside the lens until it reaches the ‘ back face of the lens. At the back face boundary, each ray of light will again refract and bend away from the normal to the surface as it moves from denser to rarer medium. The course of ray of light is shown in the following figure.
Thus, because of the concave shape of both the faces, the double concave lens diverge the rays of parallel beam of incident light.
(b) $$h_{0} = 2.0\ cm , f = -15\ cm $$
$$v = -10\ cm $$ as concave lens always forms virtual image.
Using, $$ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} $$ , we get
$$ \dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f} = \dfrac{1}{-10} - \dfrac{1}{-15} = \dfrac{-15 + 10}{150} $$
$$ \Rightarrow$$ $$u = - \dfrac{150}{5} = -30\ cm $$
Using, $$m = \dfrac{h_{f}}{h_{o}} = \dfrac{v}{u} $$ we get
$$m = \dfrac{-10}{-30} = \dfrac{1}{3} < 1 $$ we get
$$h_{f} = 2 \times \dfrac{-10}{-30} = \dfrac{2}{3} $$.
So, image is virtual , erect and diminished.
In figure $$S$$ is a monochromatic pint source emitting light of wavelength $$\lambda = 500nm$$. A thin lens of circular shape and focal length $$0.01m$$ is cut into identical halves $$L_1$$ and $$L_2$$ by a plane passing through a diameter. The two halves are placed symmetrically about the central axis $$SO$$ with a gap of $$0.5mm$$. The distance along the axis from $$S$$ to $$L_1$$ and $$L_2$$ is $$0.15m$$, while that from $$L_1$$ and $$L_2$$ to $$O$$ is $$1.30m$$. The screen at $$O$$ is normal to $$SO$$.
(i) If the third intensity maximum occurs at point $$A$$ on the screen, find the distance $$OA$$.
(i) The image of $$S$$ due to both parts of lenses are formed at $$S_1$$ and $$S_2$$ whose distance $$v$$ from the lens is given by
$$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$$
Here, $$u = -0.15m, \ f = 0.10m$$.
$$\therefore \dfrac{1}{0.10}=\dfrac{1}{v} - \dfrac{1}{(-0.15)}$$
$$\Rightarrow \dfrac{1}{v} = \dfrac{1}{0.10} - \dfrac{1}{0.15} = \dfrac{3.2}{0.30} = \dfrac{1}{0.30}$$
$$\therefore v = 0.30m$$
As the ray passing through optical center passe undeviating therefore angle $$\theta$$ subtended by lens gap and images $$S_1$$ and $$S_2$$ must be same. We, have,
$$\theta = \dfrac{O_1O_2}{u} = \dfrac{S_1S_2}{u+v}$$
$$\therefore S_1S_2 = \dfrac{u+v}{u} O_1O_2 = \dfrac{0.30+0.15}{0.15} \times 0.5mm = 1.5mm$$
i.e., $$S_1S_2 = d = 1.5mm = 1.5\times 10^{-3}m$$
Also, $$D = 1.30 - 0.30 = 1.00m$$.
$$S_1$$ and $$S_2$$ act as coherent sources and produce interference.
The positions of maximum are given by
$$y_n = \dfrac{nD\lambda}{d}$$
As point $$A$$ is third maxima,
$$\therefore y_n = OA = \dfrac{3D\lambda}{d} = \dfrac{3\times 1.00\times 500\times 10^{-9}}{1.5\times 10^{-3}}$$
$$= 1000\times 10^{-6}m = 1.0mm$$
In an experiment with a rectangular glass slab, a student observed that a ray of light incident at an angle of $$55^{\circ}$$ with the normal on one face of the slab, after refraction strikes the opposite face of the slab before emerging out into air making an angle of $$40^{\circ} $$ with the normal. Draw a labelled diagram to show the path of this ray. What value would you assign to the angle of refraction and angle of emergence?
OA - incident ray
$$i$$ is angle of incidence = $$55^{\circ} $$
Given $$r_{2} = 40^{\circ} $$
$$r_{1} $$ and $$r_{2} $$ are alternate angles,
$$ \therefore $$ $$ \angle r_{1} = \angle r_{2} = 40^{\circ} $$
So, angle of refraction = $$40^{\circ} $$
Since, the emergent ray is parallel to the incident ray, the angle of emergent must be equal to angle of incidence , i.e., $$ \angle e = \angle i = 55^{\circ} $$
$$i$$ is angle of incidence = $$55^{\circ} $$
Given $$r_{2} = 40^{\circ} $$
$$r_{1} $$ and $$r_{2} $$ are alternate angles,
$$ \therefore $$ $$ \angle r_{1} = \angle r_{2} = 40^{\circ} $$
So, angle of refraction = $$40^{\circ} $$
Since, the emergent ray is parallel to the incident ray, the angle of emergent must be equal to angle of incidence , i.e., $$ \angle e = \angle i = 55^{\circ} $$
A ray of light, incident obliquely on a face of a rectangular glass slab placed in air, emerges from the opposite face parallel to the incident ray. Draw labelled diagram to show it. State two factors on which the lateral displacement of the emergent ray depends.
Perform an experiment to demonstrate that light bends from its path, when it falls obliquely on the surface of a glass slab. Also show that angle of incidence is about equal to the emergent angle.
- Take a glass slab and place it on a white sheet of paper fixed on a drawing board.
- Mark the boundary ABCD of the glass slab.
- Fix two pins $$P_{1}$$ and $$P_{2}$$ , vertically on the drawing board such that line joining the pins is inclined to the edge AB of the glass slab (Refer Figure).
- Now, look through the glass slab from opposite side so that the images of pins $$P_{1}$$ and $$P_{2}$$ are seen exactly in line. Fix two pins $$P_{3}$$ and $$P_{4}$$, vertically on the drawing board such that pins $$P_{3}, P_{4}$$ and image of $$P_{1}$$ and $$P_{2}$$ are exactly in one line.
- Remove the glass slab and all pins. Join the points $$P_{1}$$ and $$P_{2}$$ with a line and extend this line to touch the edge AB at P. Similarly join the points $$P_{3}$$ and $$P_{4}$$ with a line and extend the line to touch the edge DC at Q.
- Join points P and Q with a straight line.
- Draw normal $$NN’$$ at P on the edge AB and a normal $$N_{1}N_{1}'$$ at Q on the edge DC.
- Measure $$\angle P_{2}PN $$ and $$\angle P_{3}QN_{1}' $$ using a protractor or dee.
- This experiment shows that when light falls obliquely on a glass slab, it bends along PQ from its original path along PL.
- Also $$\angle P_{2}PN \approx P_{3}QN_{1}'$$. That is, angle of incidence is about equal to the emergent angle.
- Note : Distance between pins $$P_{1}$$ and $$P_{2} $$, and between $$P_{3} $$ and $$P_{4} $$ must be large
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