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CBSE Questions for Class 12 Medical Physics Wave Optics Quiz 1 - MCQExams.com
CBSE
Class 12 Medical Physics
Wave Optics
Quiz 1
According to Huygens, the ether medium pervading entire universe is
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Less elastic and more dense
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Highly elastic and less dense
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Not elastic
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Much heavier
Explanation
Huygen considered, light needs a medium to propagate called ether which is highly elastic and less denser.
Which of the following properties shows that light is a transverse wave?
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Reflection
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Interference
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Diffraction
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Polarization
Explanation
The polarization phenomenon, verifies the transverse nature of light. Since sound has longitudinal nature, so it does not show polarization effect.
Two sources are called coherent if they produce waves
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of equal wavelength
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of equal velocity
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having same shape of wavefront
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having a constant phase different
Explanation
Two source are called coherent, if they have constant phase difference i.e., phase difference between two wave is constant with respect to time.
Ray optics is valid when characteristic dimensions are
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of the same order as the wavelength of light
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much smaller than the wavelength of light
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much larger than the wavelength of light
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of the order of 1 mm
Explanation
Ray optics is valid when characteristics dimensions are larger than the wavelength of the light, so that rectilinear property of light can be used.
Light travels in a ________ path
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rectilinear
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zig zag
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circular
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helical
Explanation
Light travels in straight line unless it passes through a change in medium.
Light, like sound, cannot pass through vacuum. State whether true or false
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True
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False
Explanation
Light is a form of energy which doesn't need any medium to propagate.Whereas sound travels via disturbances in the medium.
State whether true or false.
Light is a form of energy that causes a sensation of smell.
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True
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False
Explanation
Smell sensation is due to chemical present in our atmosphere.
The inability of a lens to bring all the rays coming from a point object to focus at one single point is called
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Spherical aberration
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Parallex
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Optical illusion
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none
Explanation
(A) Spherical aberration.
Light is a form of _______ that we can detect with our ________ .
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energy, ears
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corpuscles, eyes
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energy, eyes
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sensation, skin
Explanation
Light is a form of energy that we can detect with our eyes
Which of the following statements about the behaviour of light is not correct?
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Interference patterns are evident for light behaving as rays.
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Ray properties of light are useful for understanding how images are formed by optical devices such as eyes.
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Wave properties are important for observing the behaviour of light at a fine scale.
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Both wave and particle theories of light can be related to the colour sensations produced by light.
Explanation
Interference patterns are explained using wave nature of light. You can learn more from youtube video "Interference of light".
To demonstrate the phenomenon of interference, we require two sources which emit radiation of.
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Nearly the same frequency
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The same frequency
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Different wavelengths
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The same frequency and having a definite phase relationship
Explanation
As for an Interference, Two Sources must be Coherent, hence the two sources must have the same frequency and a definite phase relationship.
In a Young's double slit experiment the slit separation is $$1\ mm$$ and wavelength of light used is $$6500\ \mathring { A }$$. The distance of the seventh bright fringe from the second dark fringe formed on a screen placed $$1\ m$$ away is
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$$1.8\ mm$$
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$$3.6\ mm$$
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$$7.2\ mm$$
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$$0.9\ mm$$
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$$0.18\ mm$$
In an astronomical microscope, the focal length of the objective is made :
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shorter than that of the eye piece
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greater than that of the eye piece
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half of the eye piece
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equal to that of the eye piece
Explanation
In an Astronomical telescope, the objective lens has a greater radius than the eyepiece.
Thus the objective lens has a greater focal length than the eyepiece.
Option B is correct.
In Newton's rings experiment , light of wavelength 5890 A$$^{0}$$ is used the order of the dark ring produced where the thickness of the air film id 589 mm is
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2
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3
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4
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5
Light has a wave nature,because-
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The light travel in a straight line
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Light exhibits phenomenon of reflection and refraction
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Lights phenomeno interference
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Light exhibits phenomeno of photo electric effect
The resolving power of a telescope depends on :
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length of telescope
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focal length of objective
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diameter of the objective
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focal length of eyepiece
Explanation
Resolving power of telescope $$R=\dfrac{1}{\Delta \theta}=\dfrac{a}{1.22 \lambda}$$
where, $$\Delta \theta$$ is angular separation between two objects.
$$a$$ is the diameter of the objective.
$$\lambda$$ is wavelength of light.
So, clearly resolving power of a telescope depends on diameter of the objective.
In a room containing smoke particles, the intensity source of light will
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obey the inverse square law
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be constant at all distances
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increase with distance from the source than the inverse fourth power law
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fall faster with distance from the source than the inverse fourth power law
In a Fraunhoffer diffraction experiment at a single slit using light of wavelength $$400\ nm$$, the first minimum is formed at an angle of $$30^{\circ}$$. Then the direction $$\theta$$ of the first secondary maximum is
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$$\tan^{-1}\left (\dfrac {4}{3}\right )$$
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$$60^{\circ}$$
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$$\sin^{-1}\left (\dfrac {3}{4}\right )$$
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$$\tan^{-1}\left (\dfrac {3}{4}\right )$$
If diffraction occurs through a single slit then intensity of first secondary maxima become......% of central maxima
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4%
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25%
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75%
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50%
A man wants to see two poles, separately, situated at $$11 km$$. The minimum distance (approximately) between these poles will be
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$$5m$$
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$$2.2m$$
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$$1m$$
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$$3m$$
Interference of light from two sources can be observed if
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The sources are independent.
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The sources are of different frequencies and random phases.
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The sources are of different frequency.
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The sources are coherent.
Explanation
Interference of light from two sources can be observed if the sources are coherent . It means that the phase difference between two sources must be constant with time . Two independent sources can't produce an interference pattern .
The wave theory in its original form was first postulated by
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Issac Newton
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Thomas Young
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Christian Huygens
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Augustine Jean Fresnel
Explanation
Christian Huygens postulated
wave theory.
In the set up shown, the two slits $$S_{1}$$ and $$S_{2}$$ are not equidistant from the slit S. The central fringe at O is then
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always bright
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always dark
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either dark or bright depending on the position of S
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neither dark nor bright
Explanation
As the two slits $$S_{1}$$ and $$S_{2}$$ are not equidistant from the slit s the distance traversed by light through $$S_{1}$$ and $$S_{2}$$ may not differ by an integral multiple of wavelength. Thus it need not be bright . similarly it need not be dark. Thus option C is correct.
In studying diffraction pattern of different obstacles, the effect of:
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full wave front is studied
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portion of a wave front is studied
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waves from two coherent sources is studied
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waves from one of the coherent source is studied.
Explanation
Diffraction pattern is best explained by Huygen's wave theory in which a portion of a wavefront is studied.
Diffraction of light was discovered by :
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Young
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Hertz
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Grimaldi
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Malus
Explanation
The effects of diffraction of light were first carefully observed and characterized by Maria Grimaldi.
In diffraction pattern:
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The fringe widths are equal
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The fringe widths are not equal
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The fringes can not be produced
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The fringe width may or may not be equal
Explanation
In diffraction pattern minima are obtained at angles $$\theta _{n}$$ given by:
$$ d sin \theta _{n}=n\lambda $$
Because of sinusoidal dependence the fringe widths are not equal.
Assertion : Resolving power of a telescope is more if the diameter of the objective lens is more.
Reason : Objective lens of large diameter collects more light
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Both Assertion and Reason are correct and Reason is correct explanation of Assertion
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Assertion and Reason both are correct but Reason is not correct explanation of Assertion.
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Assertion is true but Reason is false.
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Both Assertion and Reason are false.
Explanation
Resolving power of a telescope is more if the diameter of the objective lens is more because $$R=\dfrac{a}{1.22 \lambda}$$
where, a is diameter of the objective. objective lens of large diameter collects more light but does not increase the resolving power of the telescope because resolving power increases when angular separation increases.
To demonstrate the phenomenon of interference we require two soruces which emit
radiation of
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nearly the same frequency
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the same frequency
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different wavelength
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the same frequency and having a definite phase relationship.
Explanation
To demonstrate interference, to coherent sources are required. Sources are called coherent when they emit waves of nearly equal or equal frequency and a constant phase difference throughout.
In a double-slit experiment, at a certain point on the screen the path difference between the two interfering waves is $$\cfrac { 1 }{ 8 } th$$ of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is:
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$$0.672$$
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$$0.853$$
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$$0.760$$
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$$0.568$$
Explanation
Given path difference $$\Delta x = \dfrac{\lambda}{8}$$
Phare difference $$\Delta \phi = \dfrac{2\pi}{\lambda} . \Delta x$$
$$\Delta \phi = \dfrac{2\pi }{\lambda} . \left(\dfrac{\lambda}{8}\right)$$
$$\Delta \phi = \dfrac{\pi}{4}$$
In $$YDSE$$ we know that
at any point, intensity $$I = I_{max} \cos^2\left(\dfrac{\Delta \phi}{2}\right)$$
$$I = I_{max} \cos^2\left(\dfrac{\pi/4}{2}\right)$$
$$\dfrac{I}{I_{max}} = \left[ \cos\left(\dfrac{\pi}{8}\right)\right]^2$$
$$\dfrac{I}{I_{max}} = (0.9238)^2$$
$$\dfrac{I}{I_{max}} = 0.853$$
In Young's double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit double of that from other slit. If $$\mathrm{I}_{\mathrm{m}}$$ be the maximum intensity. The resultant intensity $$I$$ when they interfere at phase difference $$\phi$$ is given by
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$$\displaystyle \frac{I_{m}}{9}(4+5\cos\phi)$$
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$$\displaystyle \frac{I_{m}}{3}(1+2\cos^{2}\frac{\phi}{2})$$
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$$\displaystyle \frac{I_{m}}{5}(1+4\cos^{2}\frac{\phi}{2})$$
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$$\displaystyle \frac{I_{m}}{9}(1+8\cos^{2}\frac{\phi}{2})$$
Explanation
$$I= I_0+ 4I_0+ 2\sqrt{I_0\times 4I_0 }cos\phi$$
$$I= I_0+ 4I_0+ 4I_0 cos\phi$$
$$I_m$$ will be at $$cos\phi=1$$,
So, $$I_m= 9I_0$$
$$\displaystyle \frac{I}{I_m}= \frac{5I_0+ 4I_0 cos\phi}{9}$$
$$\displaystyle \frac{I}{I_m}= \frac{1+ 8cos^2\frac{\phi}{2}}{9}$$
The angular width of the central maximum in a single slit diffraction pattern is $$60^o$$. The width of the slit is $$1$$ $$\mu$$m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance $$50$$cm from the slits. If the observed fringe width is $$1$$cm, what is slit separation distance? (i.e., distance between the centres of each slit.)
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$$75\mu$$m
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$$100\mu$$m
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$$25\mu$$m
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$$50\mu$$m
Explanation
Angular width of the central maxima $$\theta_c = 60^o$$
Half angle $$\theta =\dfrac{60^o}{2} = 30^o$$
Given : $$b = 10^{-6} \ m$$
Using $$\sin\theta = \dfrac{\lambda}{b}$$
Or $$\sin 30^o = \dfrac{\lambda}{10^{-6}}$$
$$\implies \ \lambda = 0.5\times 10^{-6} \ m$$
Given : $$D = 50 \ cm = 0.5 \ m$$
Fringe width $$\beta = 1 \ cm = 0.01 \ m$$
Using $$\beta = \dfrac{\lambda \ D}{d}$$
$$\therefore$$ $$0.01 = \dfrac{0.5\times 10^{-6}\times 0.5}{0.01} = 25 \mu m$$
A single slit of width b is illuminated by a coherent monochromatic light of wavelength $$\lambda$$ . If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are at $$3$$ cm and $$6$$ cm respectively from the central maximum, what is the width of the central maximum? (i.e. distance between first minimum on either side of the central maximum)
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$$4.5 cm$$
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$$6.0 cm$$
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$$1.5 cm$$
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$$3.0 cm$$
Explanation
Difference between second and first minimum $$=6-3=3$$
Which is also given as $$x_y-x_2$$
Where $$x_n = \dfrac{D}{d}\left[(2n-1)\dfrac{\lambda}{2}\right]$$
$$\therefore x_y = \dfrac{D}{d}\left(\dfrac{7\lambda}{2}\right), x_2=\dfrac{3\lambda D}{2d}$$
$$\Rightarrow x_y-x_2=3cm=\dfrac{D}{d} \left(\dfrac{7\lambda}{2}-\dfrac{3\lambda}{2}\right)$$
$$\Rightarrow 3cn = 2\lambda \dfrac{D}{d} \Rightarrow \dfrac{D\lambda}{d}=\dfrac{3}{2}=1.5cm$$
Now, width of central maximum $$= 2\times x_1$$
$$=2\times \dfrac{D}{d}\left[(2\times 1-1)\dfrac{\lambda}{2}\right]$$
$$=\dfrac{D\lambda}{d}$$
$$=1.5cm$$
For the propagation of light wave, medium is required. This is according to
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Maxwell's theory
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Huygen's theory
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Planck's theory
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Newton's theory
Explanation
Huygens suggested that light may be a wave phenomenon produced by mechanical vibrations of an all pervading hypothetical homogenous medium called eather just like those in solids and liguid .This medium was supposed to be mass less with extremely high elasticity and very low density.
The box of a pinhole camera of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength $$\lambda $$ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size say $${b }_{ min }$$ when:
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$$a=\dfrac { { \lambda }^{ 2 } }{ L } $$ and $${ b }_{ min }=\dfrac { { 2\lambda }^{ 2 } }{ L } $$
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$$a=\sqrt { \lambda L } $$ and $${ b }_{ min }=\dfrac { { 2\lambda }^{ 2 } }{ L } $$
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$$a=\sqrt { \lambda L } $$ and $${ b }_{ min }=\sqrt { 4\lambda L } $$
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$$a=\dfrac { { \lambda }^{ 2 } }{ L } $$ and $${ b }_{ min }=\sqrt { 4\lambda L } $$
Explanation
We know that spot size $$b=2(\dfrac { \lambda L }{ 2a } )+2a$$.
$${ a }^{ 2 }+\lambda L-ab=0$$.
For roots to be real D >= 0,
Hence, $${ b }_{ min }=\sqrt { 4\lambda L } $$ in which case $$a=\sqrt { \lambda L } $$
In a Young's double slit experiment, the slit separation d is $$0.3$$ mm and the screen distance D is $$1 m$$. A parallel beam of light of wavelength $$600 \, nm$$ is incident on the slits at angle $$\alpha$$ as shown in figure. On the screen, the point O is equidistant from the slits and distance PO is $$11.0 $$mm. Which of the following statement (s) is/are correct ?
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For $$\alpha = 0$$, there will be constructive interference at point P.
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For $$\alpha = \dfrac{0.36}{\pi} $$ degree, there will be destructive interference at point P.
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Fro $$\alpha = \dfrac{0.36}{\pi}$$ degree, there will be destructive interference at point O.
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Fringe spacing depends on $$\alpha$$.
Explanation
$$\Delta x = d \sin \alpha + d \sin \theta$$
$$\theta , \alpha$$ : small angle $$ \sin \theta \simeq \tan \theta = \dfrac{y}{b}$$
$$\Delta x = d \alpha + \dfrac{dy}{D}$$
$$(1) \, \alpha = 0 $$ $$\therefore \Delta x = dy/ D = \dfrac{0.3 \times 11}{1000} = 33 \times 10^{-4} mm$$
$$\Delta x$$ in terms of $$ \lambda = \dfrac{33 \times 10^{-4}}{600 \times 10^{-6}} \lambda = \dfrac{11 \lambda}{2}$$
as $$\Delta x = (2n - 1) \dfrac{\lambda}{2}$$
There will be destructive interference
$$(2) \, \Delta x = 0.3 mm \times \dfrac{0.36}{\pi} \times \dfrac{\pi}{180} + \dfrac{0.3 mm \times 11 mm}{1000} = 39 \times 10^{-4} mm$$
$$39 \times 10^{-4} = (2n - 1) \times \dfrac{600 \times 10^{-9} \times 10^3}{2}$$
$$n = 7$$
There will be destructive interference
$$(3) \, \Delta x = 3 mm \times \dfrac{0.36}{\pi} \times \dfrac{\pi}{180} + 0 = 600 nm$$
$$600 nm = n \lambda$$
$$n = 1$$
constructive interference
(4) Fringe width does not depend on $$\alpha $$.
Two slits in Young's experiment have widths in the ratio 1 :The ratio of intensity at the maxima and minima in the interference pattern, $$\displaystyle \frac{I_{max}}{I_{min}}$$ is
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$$\displaystyle \frac{4}{9}$$
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$$\displaystyle \frac{9}{4}$$
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$$\displaystyle \frac{121}{49}$$
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$$\displaystyle \frac{49}{121}$$
Explanation
Intensity is proportional to width of the slit. thus, $$\dfrac{I_1}{I_2}=\dfrac{1}{25}$$
or $$\dfrac{a_1}{a_2}=\sqrt{\dfrac{I_1}{I_2}}=\dfrac{1}{5}$$ or $$a_2=5a_1$$
now, $$\dfrac{I_{max}}{I_{min}}=\dfrac{(a_1+a_2)}{(a_1-a_2)^2}=\dfrac{(a_1+5a_1)^2}{(a_1-5a_2)^2}=36/16=9/4$$
A beam of light of $$\lambda = 600 nm$$ from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between first dark fringes on either side of the central bright fringe is
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1.2 cm
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1.2 mm
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2.4 cm
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2.4 mm
Explanation
Angular width of $$1^{st}$$ maxima
$$2\theta=\dfrac {2\lambda}{a}$$
Linear width of $$1^{st}$$ maxima $$=(D)(2\theta)=\dfrac {2\lambda D}{a}=\dfrac {2\times (600\times 10^{-9})(2)}{1\times 10^{-3}}=2.4 mm$$
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength $$5\times 10^{-5}$$ cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is:
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0.15 cm
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0.10 cm
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0.25 cm
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0.20 cm
Explanation
$$f = D = 60\ cm$$
For first minima,
$$y = \dfrac{\lambda D}{a} = \dfrac{5 \times 10^{-7} \times 60}{2 \times 10^{-2} \times 10^{-2}} = 0.15\ cm$$
For a parallel beam of monochromatic light of wavelength $$'\lambda'$$, diffraction is produced by a single slit whose width 'a' is of the order of the wavelength of the light. If 'D' is the distance of the screen from the slit, the width of the central maxima will be
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$$\dfrac {Da}{\lambda}$$
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$$\dfrac {2Da}{\lambda}$$
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$$\dfrac {2D\lambda}{a}$$
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$$\dfrac {D\lambda}{a}$$
Explanation
For minima:
$$\Delta x=\dfrac{\lambda}{2}$$...(i)
$$\dfrac{a sin\theta}{2}=\dfrac{\lambda}{2}$$...(ii)
$$sin\theta=tan\theta=\dfrac{y}{D}$$
from equation (ii) and (iii)
$$y=\dfrac{\lambda D}{a}$$
the second minima will form $$y=\dfrac{\lambda D}{a}$$ exactly below central minima
width of central maxima$$=2\dfrac{\lambda D}{a}$$
In a diffraction pattern due to a single slit of width $$'a'$$, the first minimum is observed at an angle $${30}^{o}$$ when light of wavelength $$5000\mathring { A } $$ is incident on the slit. The first secondary maximum is observed at an angle of :
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$$\sin ^{ -1 }{ \left( \dfrac { 1 }{ 4 } \right) } $$
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$$\sin ^{ -1 }{ \left( \dfrac { 2 }{ 3 } \right) } $$
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$$\sin ^{ -1 }{ \left( \dfrac { 1 }{ 2 } \right) } $$
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$$\sin ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $$
Explanation
For first minima: $$sin\ 30^0=\dfrac{\lambda}{a}=\dfrac{1}{2}$$
First secondary maxima will be at:
$$sin \ \theta=\dfrac{3\lambda}{2a}=\dfrac{3}{2}\times \dfrac{1}{2}$$
$$\Rightarrow \theta=sin^{-1}\left ( \dfrac{3}{4}\right )$$
In Youngs double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :
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half
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four times
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one-fourth
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double
Explanation
Fringe width $$ \beta =\dfrac { \lambda D }{ d } $$
On $$ d'=\dfrac { d }{ 2 } ,D'=2D$$
New fringe width $$\beta '=\dfrac { \lambda D' }{ d' } =4\beta $$
In a single slit diffraction with $$\displaystyle \lambda =500nm$$ and a lens of diameter 0.1 mm, width of central maxima, obtain on screen at a distance of 1 m will be
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$$5 mm$$
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$$1 mm$$
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$$10 mm$$
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$$2.5 mm$$
Explanation
Angle subtended by two minima at the slit in a single slit diffraction $$=\alpha=\dfrac{2\lambda}{w}$$
where, $$w$$ is the slit width.
Here the lens' diameter would act as slit width.
The width of central maxima is the distance between the two minima $$=d\alpha$$
where, $$d$$ is the distance between slit and screen $$=1\ m$$
Thus, the width of central maxima $$=10\ mm$$
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
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Assertion is correct but Reason is incorrect.
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Both Assertion and Reason are incorrect.
Explanation
A laser is a highly monochromatic and near perfect parallel beam of light, due to which the beam can be focussed by a converging lens to a very small spot. As the intensity of the beam is too high, it can drill holes through a metal sheet even if the power is $$0.2\ W$$. But even a torch-light of $$1000\ W$$ power cannot drill holes in such a metal sheet, because the light is less intense and the beam is not parallel.
Therefore, assertion is correct but reason is incorrect.
Resolving power of a telescope increases with :
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increase in focal length of eyepiece
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increase in focal length of objective
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increase in aperture of eyepiece
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increase in aperture of objective
Explanation
Resolving power of a telescope:
$$R=\dfrac{a}{1.22 \lambda}$$
where, $$a$$ is diameter of the objective
so, $$R$$ increases when a is increased and $$a$$ increases when aperture of objective is increased
A beam of light of wavelength $$600 \ nm$$ from a distant source falls on a single slit $$1.00 \ mm$$ wide and the resulting diffraction pattern is observed on a screen $$2 \ m$$ away. The distance between the first dark fringe on either side of the central maxima is :
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$$1.2 \ cm$$
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$$1.2\ mm$$
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$$2.4 \ mm$$
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$$4.8\ mm$$
Explanation
Condition for obtaining $$m^{th}$$ order minima is
$$a sin\theta=m\lambda$$
Thus for obtaining first diffraction minima,
$$ a sin\theta=\lambda$$
$$\implies sin\theta=\dfrac{600\times 10^{-9}m}{10^{-3}m}\approx tan\theta=\dfrac{y}{D}$$
$$\implies y=1.2\ mm$$
Thus distance between fringes on either side=$$2\times y=2.4\ mm$$
A: In interference pattern, intensity of successive fringes due to achromatic light is not same.
R: In interference, only redistribution of energy takes place.
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Both A and R are true, and R is correct explanation of A
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Both A and R are true, and R is not correct explanation of A
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A is true but R is false
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A is false but R is true
Explanation
In Interference, the resultant pattern is determined by the path difference between the two waves. It is same for monochromatic as well as achromatic light. Hence statement A is false. Statement R is true because in interference only redistribution of energy takes place.
A wavefront is an imaginary surface where :
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phase is same for all points
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phase changes at constant rate at all points along the surface
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constant phase difference continuously changes between the points
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phase changes all over the surface
Explanation
$${\textbf{Explanation:}}$$
$$\bullet$$
The locus of all particles in a medium, vibrating in the same phase is called wave front.
$$\bullet$$
The direction of propagation of light (ray of light) is perpendicular to the wave front.
$$\bullet$$
A wave front is an imaginary surface where all particles lying on this vibrate in the same phase.
$${\textbf{Correct option: A}}$$
When two light waves meet at a place :
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their displacements add up
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their intensities add up
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both will add up
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energy becomes zero
Explanation
Superposition means when two light waves meet at a place their resultant displacement is the sum of initial displacements.
A : The phase difference between any two points on a wave front is zero
R : From the source light, reaches every point on the wave front in the same time.
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Both A and R are true, and R is correct explanation of A
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Both A and R are true, and R is not correct explanation of A
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A is true but R is false
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A is false but R is true
Explanation
A wave front is the locus of points having the same phase.
Thus statement A is true because the phase difference is zero as all points have same phase.
Statement R is true because a wavefront is composed of all points where the light reaches from the source in the same time.
As light takes the same time it has the same phase at all points on a wavefront and hence it correctly explains statement A.
Huygens' wave theory is used :
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to determine the velocity of light
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to find the position of the wave front
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to determine the wavelength of light
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to find the focal length of a lens
Explanation
Huygen proposed a hypothesis for the geometrical construction of the position of a common wavefront at any instant during the propogation of waves in a medium.
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Practice Class 12 Medical Physics Quiz Questions and Answers
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