Light Reflection And Refraction - Class 10 Physics - Extra Questions

Angle of reflection is the angle between reflected ray and the ........(incident ray, normal).

Angle of reflection is the angle between reflected ray and the normal.

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Define focal length of a spherical lens.

 Focal length of a lens is the distance between centre of lens and focal point . Focal point is a point on principal axis , at which light rays parallel to principal axis meet after refraction through lens or seem to meet after refraction .


What is reflection? Give two examples.

A phenomenon of returning light from the surface of an object when the light is incident on it is called reflection of light.
Examples:
  • Reflection by a plane mirror.
  • Reflection by a spherical mirror.

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Define normal.

Normal - The perpendicular drawn to the reflecting surface is called normal at that point.
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The process of sending back the light rays into the same medium is called _________.

The process of sending back the light-rays into the same medium is called Reflection of light.


State the characteristics of the image formed by a plane mirror.

(i) Plane mirror forms an erect image.
(ii) It forms a virtual image.
(iii) Size of the image is same as that of the object.
(iv) Image is formed at the same distance behind the mirror as the object stands in front of it.
(v)Image formed is a laterally inverted image i.e., right hand side of the object seems to be the left hand side and vice-versa.


What is reflection of light ? Also write the laws of reflection.

The phenomenon of bouncing back of light rays in the same medium on striking the smooth surface is known as reflection of light.
There are two laws of reflection :
1) Angle of incidence is always equal to angle of reflection , i.e., $$\angle i = \angle r$$.
2) The incident ray, the reflected ray and the normal at the point of incidence, all lie in the same plane.


Angle of reflection is the angle between ________ and the ________.

The angle of reflection is the angle between the reflected ray and the normal.
In the above image, $$r$ is the angle of reflection.

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The magnification produced by a mirror is - 1.What does it signify about the image formed ?

Magnification is the ratio of image size to object size.
         $$M=\pm \dfrac {h'}h$$         ($$\pm$$ according to sign convention)
When the magnification is $$-1.5$$, it implies that image size is $$1.5$$ times than the actual object size but is inverted.


  • The incident ray, refracted ray, and the normal to the interface of any two given media, all lie in the same plane
  • This is a part of the laws of refraction of light. Type $$1$$ for true and $$0$$ for false.

Law of refraction states that:

  • The incident ray, reflected ray and the normal, to the interface of any two given mediums; all lie in the same plane.
  • The ratio of the sine of the angle of incidence and sine of the angle of refraction is constant.

 


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In refraction through a glass slab with parallel faces, incident ray and emergent ray are parallel. Type 1 for true and 0 for false.

Using Snell’s law or law of refraction, we can predict the amount of bending. It is used to understand the relation between angle of incidence and refraction.

The extent of bending of the ray of light at the opposite parallel faces (air-glass interface) and (glass-air interface) of the rectangular glass slab is equal and opposite. This is why the ray emerges parallel to the incident ray.

We can also prove this geometrically, because $$\angle2$$ and $$\angle3$$ are alternate interior angles.
So, $$\angle2=\angle3$$

By Snell's law at first surface 
$$n_{glass}=\dfrac{sin \ 1}{sin \ 2}$$

If light light was incident at second surface, then the path of light would be still the same.
So by Snell's law at second surface 
$$n_{glass}=\dfrac{sin \ 4}{sin \ 3}$$

$$\because \angle2=\angle 3 \implies \angle 1= \angle 4$$ 
Therefore, incident and emergent rays are parallel.





Snell's law of refraction is $$\mu_{1}\sin\theta_{1} = \mu_{2}\sin\theta_{2}$$.Type 1 for true and 0 for false.

Snell's law of refraction is $$\mu_{1}\sin\theta_{1} = \mu_{2}\sin\theta_{2}$$. Hence, given statement is true.
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Write laws of refraction of light. Explain the same with the help of ray diagram, when a ray of light passes through a rectangular glass slab.


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When a ray of light passes from a rarer to denser medium it bends towards the normal. (True=1 or false=0).

Laws of refraction;-
  1. The incident ray,the refracted ray and the normal to the refracting surface at the point of incidence lie in the same plane.
  2. For a given pair of media and for a given colour of light the ration between the sine of angle of incidence to the sine of refraction is a constant.This constant is known as refractive index of the second medium with respect to the first medium.
When a ray of light passes through a glass slab, $$\angle i, \angle r$$ and the normal all lie in the same plane.
When a ray of light passes from one medium to another, here from air to glass or glass to air, the ratio $$\frac{sin i}{sin r}=constant$$.
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 In snells law $$\mu_{1}\sin\theta_{1} = \mu_{2}\sin\theta_{2}$$
Type 1 for true and 0 for flase

Snell's law is a formula used to describe the relationship between the angles of incidence and refraction, when referring to light or other waves passing through a boundary between two different isotropic media, such as water, glass or air.

In optics, the law is used in ray tracing to compute the angles if incidence or refraction, and in experimental optics to find the refractive index of a material
Snell's law states that the ratio of the sines of the angles of incidence and refraction is equivalent to the ratio of phase velocities in the two media, or equivalent to the reciprocal of the ratio of the indices of refraction:

$$\dfrac{sin\, \theta_2}{sin\, \theta_1}=\dfrac{v_2}{v_1}=\dfrac{\mu_1}{\mu_2}$$

So, $$\mu_1\, sin \, \theta_1=\mu_2 \, sin\, \theta_2$$

with each $$\theta$$ as the angle measured from the normal of the boundary, $$v$$ as the velocity of light in the respective medium and $$\mu$$ as the refractive index of the respective medium.










Light rays that are parallel to principal axis of concave mirror get converged at a particular point. It is called principal focus.Type 1 for true and 0 for flase

Light rays that are parallel to principal axis of concave mirror get converged at a particular point. It is called principal focus.


........ is an imaginary line perpendicular to the surface where the reflection occurs.

Normal is an imaginary line perpendicular to the surface to where the reflection occurs.


...... ray is the ray of light reflected from a surface.

Answer REFLECTED

A light ray striking a surface is called incident ray and the ray that reflects from the surface is called reflected ray.


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Angle of incidence is the angle between ....... and the .......

Angle of incidence is the angle between incident ray and the normal line.


State the laws of refraction of light.

  1. Incident ray, reflected ray, refracted ray and the normal of the system lie in the same plane.
  2. Snells Law: $$\frac{sin\ i}{sin\ r}=\frac{n_2}{n_1}=\frac{v_1}{v_2}$$


When the convex lens acts as a magnifying glass, the object is between the .......... and the optical center of the lens.

Principal focus
When the object is placed between the principal focus and the pole, the rays tend to diverge through the normally converging lens, forming a magnified, erect and virtual image of the object.

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State whether the given statement is True or False.

The distance between the principal focus and centre of curvature of a lens is called its radius of curvature.

Radius of curvature is the distance between the pole and center of curvature.
So the given statement is false.


Match the statements in Column A, with those in Column B.

A. A lens is used as a magnifying glass is a convex lens

B. A point within the lens, such that any ray passing through it does not suffer refraction is the optical centre.

C. The diameter of a lens is aperture.

D. A lens having curved surfaces, such that it is tapering in the middle and thicker at the edges concave lens.

E. A lens having one curved and plane surface, such that it is thicker in the middle and tapering at the edges is the plano-convex lens.


The diagram in a Fig. shows a point object $$P$$ in front of a plane mirror $$MM_1$$.
$$(a)$$ Complete the diagram by taking two rays from the point $$P$$ to show the formation of its image.
$$(b)$$ In the diagram, mark the position of a eye to see the image.
$$(c)$$ Is the image formed real or virtual ? Explain why ?

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(a)The above figure shows the complete diagram.
(b)The position of eye is marked at P.
(c)Image formed is virtual.Because the light rays appeared to be coming from that point and they don't actually intersect.

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Answer the following:
If you are given a part of the hollow spherical glass, how will you convert it into a concave mirror?

The outer side of the hollow spherical glass will be polished to get a concave mirror.


List three objects that contain lenses.

Cameras, telescopes, movie projectors, eye glasses, magnifying lenses, microscopes.


For each pair of terms, explain how can  you differentiate between them.
Convex lens and concave lens

In the convex lens, the curve is outward facing, whereas, in the concave lens, the curve faces inward. When the light rays pass through the convex lens, it converges the light rays and focuses on one point. On the other hand, when the light rays go through the concave lens, it diverges the beams, i.e. they spread out.
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A light ray falls on the air glass interface at an angle of $$30^{\circ}$$. What is the angle of refraction for this rays ?
$$(n_{ag}\, =\, \displaystyle \frac {3}{2})$$.

Given ,
$${ n }_{ ag }=\dfrac { 3 }{ 2 } \quad \quad \quad \quad \angle i={ 30 }^{ 0 }$$
From snell's law 

$$n=\dfrac { sini }{ sinr } $$

By putting values of n, i in equation 
we get
$$\dfrac { 3 }{ 2 } =\dfrac { sin{ 30 }^{ 0 } }{ sinr } $$

$$\Rightarrow sinr=\dfrac { 1 }{ 3 } \\ r={ 19 }^{ 0 }{47'}{ \simeq 20 }^{ 0 }$$
the angle of refraction is 20 degree.


What is the relation between angle of incidence i and angle of refraction r in the glass ?

From snell's law 
Snell's law (also known as the snell-descartes law and the law of refraction) is a formula used to describe the relationship between the angle of incidence and refraction, when referring to light or other waves passing through a boundary between two different isotropic media, such as water, glass, or air.
$$refractive\quad index=\frac { sini }{ sinr } \\ \mu =\frac { sini }{ sinr } $$


State Snell's law of refraction of light.

Snell's law states that the relationship between angle of incidences and refraction for a wave impinging on an interface between two media with different incidences of angle is always constant, The law follows from the boundary condition that a wave be continuous across a boundary, which requires that the phase of the wave be constant on any given plane.



Figures (a) and (b) show the refraction of a ray in air incident at $$60^o$$ with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in a glass when the angle of incidence in water is $$45^o$$ with the normal to a water-glass interface [Fig.(c)].
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Refractive index of glass $$\mu_g=sin60^o/sin35^o=1.51$$
For water $$\mu_w=sin60^o/sin47^o=1.184$$
So, refractive index of glass wrt to water can be $$\mu_r=\mu_g/\mu_w=1.275$$
now, 
$$\mu_r=sin45^o/sinr=1.275\Rightarrow r=36.68^o\approx 37^o$$


(a) The refractive index of glass is 1.What is the speed of light in glass? (Speed of light in vacuum is $$3.0\times 10^{8} m s^{-1})$$
(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?

(a) Speed of light in glass is $$c/\mu=2\times 10^8m/s$$
(b) The speed of light in glass is not independent of the colour of light. The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.


Define radius of curvature.

The radius of curvature is the radius of hollow sphere of which the mirror/spherical mirror is a part of.


Define Principle focus of a plane mirror.

Principal focus - When rays of light parallel to the principal axis are
incident on a spherical mirror all the rays after reflection either converge 
to a point (in concave mirror) or appear to diverge from a point (in
convex mirror) on the principle axis This point is called the principle
focus of the mirror


State four points of differences between a Concave mirror and a Convex mirror.

Concave MirrorConvex Mirror
(i) Concave or bent-in surface                         
is the reflecting surface		(i) Outer bulging surface is the 
reflecting surface
(ii) It is a converging mirror(ii) It is a diverging mirror
(iii)It has a real focus(iii) It has a virtual focus
(iv) Image may be real or virtual(iv) Image is always virtual


An object $$5.0 cm$$ in length is placed at a distance of $$20 cm$$ in front of a convex mirror of radius of curvature $$30 cm$$. Find the position of the image, its nature, and its size.

Given: $$u=-20\ cm$$
            $$f=R/2=15\ cm$$
            Size of image, $$h=5\ cm$$
Let size of image be $$h'$$
From mirror formula, $$\cfrac{1}{u}+\cfrac{1}{v}=\cfrac{1}{f}$$
                                   $$\cfrac{1}{-20}+\cfrac{1}{v}=\cfrac{1}{15}$$
                                   $$v=\cfrac{60}{7}=8.57\ cm$$

The image is formed $$8.57 cm$$ behind the mirror. It is virtual and erect.

Magnification, $$m=-v/u=8.57/20=0.428$$
Also, $$m=h'/h=h'/5$$

From above, $$h'/5=0.428$$
                      $$h'=2.14\ cm$$


An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Given: $$f=15$$
            $$u=-10$$
From mirror formula, $$\cfrac{1}{u}+\cfrac{1}{v}=\cfrac{1}{f}$$
                                   $$\cfrac{1}{-10}+\cfrac{1}{v}=\cfrac{1}{15}$$
                                   $$v=6\ cm$$
Image is formed 6 cm to the right of the mirror. Image formed is virtual and erect.


A concave lens of focal length 15cm forms an image 10 cm from the lens . How far is the object placed from the lens? Draw the ray diagram.

Given: $$f=-15$$
            $$v=-10$$
From mirror formula, $$\cfrac{1}{v}-\cfrac{1}{u}=\cfrac{1}{f}$$
                                    $$\cfrac{1}{-10}-\cfrac{1}{u}=\cfrac{1}{-15}$$
                                    $$\cfrac{1}{u}=\cfrac{1}{15}-\cfrac{1}{10}$$
                                    $$u=-30\ cm$$

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An object of size $$7.0 cm$$ is placed at $$27cm$$ in front of a concave mirror of focal length $$18 cm$$. At what distance from the mirror should a screen be placed so that a sharp focused image can be obtained? Find the size and nature of the image.

Given:
size of object, $$o=+7 cm$$
distance of object ,$$u =-27 cm$$
focal length of concave mirror, $$f=-18 cm$$
let us take size of image=$$I$$
so, mirror formula is 
$$\dfrac{1}{v} +\dfrac{1}{u} =\dfrac{1}{f}$$
so, putting values of u and v,
$$\dfrac{1}{v} =\dfrac{1}{-18} -\dfrac{1}{-27}$$
$$\dfrac{1}{v}=\dfrac{1}{-54}$$
$$v=-54cm$$
so, image is formed on object side only.
magnification,
$$m=\dfrac{-v}{u} =\dfrac{I}{o}$$
$$m=-\dfrac{-54}{-27} =\dfrac{I}{7}$$
$$I=-14 cm$$
so, image is double in size to that of object.
Nature: real, inverted and magnified image.



Why do stars twinkle?

$$\textbf{Explanation}$$
$$\bullet$$The earth atmosphere has different layers which are not rigid which means the layers are moving. Hence, the air index is not fixed. 
$$\bullet$$The light from stars passes through the atmosphere., due to the movement of different layers, the light experience different refractive indexes at different times. 
$$\bullet$$This causes the stars to twinkle when seen from the earth.


Explain why the planets do not twinkle.

Hint: The planets are very close to earth in comparison to stars.

Part 1: Size of appearing,
$$\bullet$$The planets are very close to the earth comparative to the stars. 
$$\bullet$$Therefore the planets do not look like just a point in a the sky while they appear as a tiny disc of light while the stars appear as a point.

Part 2:Conclusion,
$$\bullet$$As the stars appear as the points in the sky, for very small angular displacement in the angular change in their image to the observer due to the variable air index makes them twinkling while the planets are tiny discs, hence they do not show the shift in their image so easily. 
$$\bullet$$Therefore planets show stability in their image and they do not twinkle.


(a) State the laws of refraction of light. Give an expression to relate the absolute refractive index of a medium with speed of light in vacuum.
(b) The refractive indices of water and glass with respect to air are 4/3 and 3/2 respectively. If the speed of light in glass is $$2 \times 10^8 \,\, ms^{-1}$$, find the speed of light in (i) air, (ii) water.

(a) Law of Refraction : (1) The ratio of sine of angle of incidence to the sine of angle of refraction for a particular pair of media is constant. Therefore, if 

angle of incidence = $$i$$ 

angle of refraction = $$r$$ 

$$\dfrac{sin i}{sin r}=constant$$

The value of this constant depends upon temperature and wavelength of light. This is also known as Snell's law. 

(2) The incident ray, the refracted ray and the normal at the point of incidence, all the lie in the same plane. 

Refractive index (of a medium)  = speed of light in air (or vaccum) / speed of light in medium 

$$\implies \mu_m=\dfrac{c}{v_m}$$

(b) $$\mu_{glass}=\dfrac{c}{v_{glass}}$$

$$\implies \dfrac{3}{2}=\dfrac{c}{2\times 10^8m/s}$$

(i)$$\implies c=3\times 10^8m/s$$

(ii) $$\mu_{water}=\dfrac{c}{v_{water}}$$

$$\implies v_{water}=\dfrac{3\times 10^8m/s}{4/3}$$

$$=\dfrac{9}{4}\times 10^8m/s$$



"A ray of light incident on a rectangular glass slab immersed in any medium emerges parallel to itself." Draw labelled ray diagram to justify the statement.

For understanding the refraction of light through a glass slab let us consider the above figure consider the figure,


Here AO is the light ray travelling in air and incident on glass slab at point O.

When ray enters the glass medium it bends towards the normal NN’ that is light ray AO gets refracted on entering the glass medium.

When ray gets refracted it now travels through the glass slab and at point B it comes out of the glass slab.

As ray OB goes from glass medium to air it again gets refracted and bends away from normal N1N'1 and goes in direction BC.

Here AO is the incident ray and BC is the emergent ray and both are parallel to each other and OB is the refracted ray.

Emergent ray BC is parallel to incident ray AO because the extent of bending of the ray of light at the opposite parallel faces which are PQ and SR of the rectangular glass slab is equal and opposite.

In the figure i is the angle of incidence, r is the angle of refraction and e is the angle of emergence.

Angle of incidence and angle of emergence are equal as emergent ray and incident ray are parallel to each other.

When a light ray is incident normally to the interface of two media then there is no bending of light ray and it goes straight through the medium.


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"The magnification produced by a spherical mirror is - $$3$$". List four informations you obtain from this statement about the mirror/image.

We know that linear magnification of a convex mirror is always +ive , the given magnification is -ive so 
(1) It is a concave mirror
(2) Image is inverted 
(3) Image is real
(4)As magnification $$=-3$$ ,which is greater than $$1$$ therefore image is enlarged .


While performing the experiment on tracing the path of a ray of light through a rectangular glass slab, in which of the following experimental setups, students likely to get the best results?
$$P_1$$ and $$P_2$$ are the positions of pins fixed by him :

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The pins must not be too close to each other, otherwise, it would be hard to align the light passing through both of them when seen from the other side of the slab. Hence $$I$$ and $$IV$$ are rejected.

Also, the angle of incidence and the distance of the point of incidence from the mid slab must be optimum for observing the shifted path of light after emergence from the other side of the slab. So, $$II$$ is rejected.

Hence, the correct answer is option $$III$$.



The absolute refractive index of glass and water are $$3/2$$ and $$4/3$$, respectively. If the speed of light in glass is $$2 \times 10^8$$ m/s, calculate the speed of light in :
(a) vacuum
(b) water

Given :    $$n_{water} = \dfrac{4}{3}$$           $$n_{glass} = \dfrac{3}{2}$$             $$v_{glass} = 2\times 10^8$$  m/s            $$n_{air} = 1$$
Speed of light in a medium        $$v_{medium} = \dfrac{c}{n}$$         
where  $$c$$ and $$n$$ are the speed of light in vacuum and refractive index of medium respectively.

(a) :     $$v_{glass} = \dfrac{c}{n_{glass}}$$

$$\therefore$$     $$2\times 10^8  = \dfrac{c}{(3/2)}$$                        $$\implies c  = 3\times 10^8$$  m/s

(b) :     $$v_{water} = \dfrac{c}{n_{water}}$$

$$\therefore$$     $$v_{water}  = \dfrac{3\times 10^8}{(4/3)}  = 2.26\times 10^8$$  m/s            


For the same value of angle of incidence, the angles of refraction in three media $$A$$, $$B$$ and $$C$$ are $${15}^{o}$$, $${25}^{o}$$ and $${35}^{o}$$ respectively. In which medium would the velocity of light be minimum?

Let the refractive index of the medium be $$n$$ and speed of light in air is  $$c$$.
Using Snell's law of refraction :       $$n_{air} \times sini = n_{medium} \times sinr$$                          where  $$n_{air} =1$$
$$\therefore $$   $$1 \times sin i = n \times sin r$$           $$\implies n = \dfrac{sin i}{sin r}$$

Velocity of light in a medium        $$v = \dfrac{c}{n} = \dfrac{c \times sin r}{sin i}$$                          $$\implies v \propto  sin$$ $$ r$$
Thus velocity of light is minimum in medium A.


If you take a photograph of your image in a plane mirror, how many meters away should you set your focus if you are $$2$$m in front of the mirror?

Image formed by the plane mirror is always virtual, upright is at the same distance from the mirror as is the object and is same shape and size as that of the object.
Total distance between object and distance = 2+2 = 4m
Since image is 4 m away from object so we should set the