Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect but Reason is correct

If $$z=\cfrac{\lambda D}{4d}$$

$${ \left[ 3+2\sqrt { 2 } \right] }^{ 2 }$$

$${ \left[ 3-2\sqrt { 2 } \right] }^{ 2 }$$

$${ \left[ 3+\sqrt { 2 } \right] }^{ 2 }$$

$${ \left[ 3-\sqrt { 2 } \right] }^{ 2 }$$

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect Reason is correct

MCQ Questions for Class 12 Medical Physics Wave Optics Quiz 13

In the interference of waves from two sources of intensities

$I_{0}$ and

$4I_{0}$ . The intensity at a point where the phase difference is

$\pi$ is

0%
$I_{0}$
0%
$2I_{0}$
0%
$3I_{0}$
0%
$4I_{0}$

In a Young's double slit experiment, the distance between the two identical slits is 6.1 times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single slit diffraction pattern is :

0%
3
0%
6
0%
12
0%
24

Explanation

$\textbf{Hint:}$

For single slit diffraction if the wavelength of light is $\lambda$ , the distance between slit and screen is D and width of the slit is d, then the width of central bright fringe is given by,

$W=\dfrac{2\lambda D}{d}$

For double-slit interference if the wavelength of light is $\lambda$ , the distance between slits and screen is D and distance between slits is d, then the width of central bright fringe is given by,

$W=\dfrac{\lambda D}{d}$

$\textbf{Step1:}$ Calculate width of central maxima for diffraction.

Let wavelength of light is $\lambda$ , width of slit is $a$ .If distance between slits and screen is $D$ , then width of maxima in interference is

${{W}_{1}} =\dfrac{2\lambda D}{ a}$

$\textbf{Step 2:}$ Calculate the width of maxima for interference.

Let wavelength of light is $\lambda$ , width of slit is $a$ then distance between slits is $6.1 \times a$ .If the distance between slits and screen is $D$ , then the width of maxima in interference is

${{W}_{2}} =\dfrac{\lambda D}{6.1\times a}$

$\textbf{Step 3}$ Find the number of maxima of double-slit interference that will lie within the central maxima of the single-slit diffraction.

Let us assume that the width of $n$ number of maxima of double-slit interference is equal to the width of central maxima of the single-slit diffraction pattern.

Then,

$n= \dfrac{{{W}_{1}}}{{{W}_{2}}}$

$\Rightarrow n=\dfrac{\dfrac{2\lambda D}{a}}{\dfrac{\lambda D}{6.1\times a}}$

$\Rightarrow n=12.2$

Since $n$ must be a whole number,

Therefore, $n=12$ .

Thus, the number of maxima of double-slit interference that will be observed within the width of central maxima of single-slit diffraction is $n=12$ .

True & False Statement Type

$S_1$ : In an elastic collision initial and final K.E. of system will be same.

$S_2$ : In a pure L-C Circuit average energy stored in capacitor is zero.

$S_3$ : In YDSE coherent sources are formed by division of wave front method.

$S_4$ : If a physical Quantity is quantized then it must be integral multiple of its lowest value.

0%
FFTF
0%
TTFT
0%
FTFT
0%
TFTT

A single slit of width

$a$ is illuminated by violet light of wave length

$400\ nm$ and width of the diffraction pattern is measured as

$y$ . Half of the slit is covered and illuminated with

$600\ nm$ . The width of the diffraction pattern will be

0%
$\dfrac{y}{3}$
0%
pattern vanishes and width is zero
0%
$3y$
0%
none of these

Explanation

We have width of the diffraction pattern

$\beta = \dfrac{2 \lambda D}{d}$
Let us consider

$\beta$ as

$y$ , then we have

$y = \dfrac{2 \lambda D}{d}$
In the first instance

$\lambda = 400 nm$

$\implies y = \dfrac{2\times 400 D}{d} nm$
In the second instance

$\lambda = 600 nm$ and

$d = \dfrac {d}{2}$

$\implies y' = \dfrac{2\times 600 D}{\dfrac {d}{2}} nm$
Hence

$\dfrac {y}{y'} = \dfrac {1}{3}$

$\implies y' = 3 y$

Find the nature and order of the interference at O

0%
$20^{th}$ minima
0%
$20^{th}$ maxima
0%
$10^{th}$ maxima
0%
$10^{th}$ minima

Explanation

The optical path difference (

$\Delta x$ ) between the waves at

$P$ is given by

$\Delta x = S_2P - S_1P = {\dfrac {y d}{D}}$
Therefore,

$\Delta x = 0.01 mm$
Also,

$\Delta x = n \lambda = 0.01 mm$

$\implies n = \dfrac {0.01 \times 10^{-3}} {5000 \times 10^{-10}} = 20$
Hence a maxima of

$20^{th}$ order is obtained at 0.

An astronomical telescope, consists of two thin lenses set

$36 cm$ a part and has a magnifying power

$8$ . Calculate the focal length of the lenses.

0%
$32 cm$
0%
$18 cm$
0%
$25 cm$
0%
$36 cm$

Explanation

Here,

$f_o+f_e=36 cm$

$M=-8$ (magnifying power is negative)
Now,

$M=\cfrac {f_o}{f_e}$

$\therefore -8=-8=-\cfrac {f_o}{f_e}$

$\Rightarrow f_o=8f_e$

From the equations, we have

$8f_e+f_e=36$ or

$f_e=4\ cm$

Again,

$f_o=8f_e=8\times 4=32\ cm$

In an experiment of single slit diffraction pattern, first minimum for red light coincides with first maximum of some other wavelength. If wavelength of red light is

$6000 A^o$ , then wavelength of first maximum will be

0%
$3000 A^o$
0%
$4000 A^o$
0%
$5000 A^o$
0%
$6000 A^o$

Explanation

First Minima of red light coincides with maxima of some other light .

$(n+\dfrac{1}{2})\lambda_{unknown} = n \lambda_{red}$

$\lambda_{unknown} = \dfrac{2}{3}\times 6000$

$\lambda_{unknown} = 4000 A^o$

Light of wavelength 589.3 nm is incident normally on a slit of width 0.1 mm. The angular width of the central diffraction maximum at a distance of 1m from the slit is

0%
$0.68^\circ$
0%
$0.34^\circ$
0%
$2.05^\circ$
0%
None of these

Explanation

Let D be the distance between source and screed d the distance between coherent source then for central diffraction maxima,
where

$\lambda$ is wavelength
Given,

$\lambda =589.3 nm=589.3\times 10^{-9} m$

$d=0.1 \times 10^{-3} m, D=1m$

$\therefore W=\dfrac{2\times 589.3 \times 10^{-9}}{0.1 \times 10^{-3}} \times 1$
Angular width

$\theta=\dfrac{W}{D}=\dfrac{2\times 589.3 \times 10^{-9}}{0.1 \times 10^{-3}} rad =0.68^{\circ}$

Find the nature and order of the interference at the point P :

0%
$70^{th}$ maxima
0%
$80^{th}$ minima
0%
$60^{th}$ maxima
0%
$70^{th}$ minima

Explanation

The optical path difference (

$\Delta x$ ) between the waves at

$P$ is given by

$\Delta x = S_2P - S_1P = {\dfrac {y_1 d}{D_1}} + {\dfrac {y_1 d}{D_1}}$

$\Delta x = \dfrac {(1)(10)}{10^3} + \dfrac {(5)(10)}{2 \times 10^3}$
Therefore,

$\Delta x = 0.035 mm$
Also,

$\Delta x = n \lambda = 0.035 mm$

$\implies n = \dfrac {0.035 \times 10^{-3}} {5000 \times 10^{-10}} = 70$
Hence a maxima of

$70^{th}$ order is obtained at P.

${M}_{1}$ and

${M}_{2}$ are plane mirrors and kept parallel to each other. At point O, there will be a maxima for wavelength

$\lambda$ . Light from a monochromatic source

$S$ of wavelength

$\lambda$ is not reaching directly on the screen. Then,

$\lambda$ is:

0%
$\cfrac { { 3d }^{ 2 } }{ D }$
0%
$\cfrac { { 3d }^{ 2 } }{ 2D }$
0%
$\cfrac { { d }^{ 2 } }{ D }$
0%
$\cfrac { { 2d }^{ 2 } }{ D }$

Explanation

Light emerging from point S gets reflected from mirrors

$M_{1}$ and

$M_{2}$ , to interfere constructively at point O.

This means that path difference between light travelling

$SM_{1}O$ and

$SM_{2}O$ must be

$\lambda$ .

As shown in the figure, path

$SM_{1}O = 2\sqrt{(\dfrac{D}{2})^2 +(\dfrac{d}{2})^2}$

path

$SM_{2}O = 2\sqrt{(\dfrac{D}{2})^2 + {d}^2}$

$2\sqrt{(\dfrac{D}{2})^2 + {d}^2}$

$-$

$2\sqrt{(\dfrac{D}{2})^2 +(\dfrac{d}{2})^2} = \lambda$

Thus using Bionomial expansion,

$\lambda$ comes out to be

$\dfrac{3d^2}{2D}$ .

The intensity of light from a source is 500 /

$\pi W/m^{2}$ . Find the amplitude of electric field in this wave -

0%
$\sqrt{3}\times 10^{2}N/C$
0%
$2\sqrt{3}\times 10^{2}N/C$
0%
$\frac{\sqrt{3}}{2}\times 10^{2}N/C$
0%
$2\sqrt{3}\times 10^{1}N/C$

Two identical coherent sources are placed on a diameter of a circle of radius

$R$ at separation

$x(<<R)$ symmetrical about the center of the circle. The sources emit identical wavelength

$\lambda$ each. The number of points on the circle of maximum intensity is

$(x=5\lambda$ ):

0%
$20$
0%
$22$
0%
$24$
0%
$26$

Explanation

For maxima:

$d\sin { \theta } =n\lambda$

On circle, for

$-\dfrac { \pi }{ 2 } \le \theta \le \dfrac { \pi }{ 2 } \Rightarrow n=\dfrac { 2d }{ \lambda } +1$

for

$\dfrac { \pi }{ 2 } <\theta <-\dfrac { \pi }{ 2 } \Rightarrow n=\dfrac { 2d }{ \lambda } -1$

Total number of maxima

$n=\dfrac { 4d }{ \lambda } =\dfrac { 4\left( 5\lambda \right) }{ \lambda } =20$

Light of wavelength

$500nm$ goes through a pinhole of

$0.2mm$ and falls on a wall at a distance of

$2m$ . What is the radius of the central bright spot formed on the wall?

0%
$2.37 cm$
0%
$1.37 cm$
0%
$3.37 cm$
0%
$7.37 cm$

Explanation

$R=\displaystyle\frac{1.22\lambda D}{r}$

$=\displaystyle\frac{1.22\times 560\times 2\times 10^{-9}}{0.1\times 10^{-9}}$

$=1.37cm$

Two points sources separated by

$2.0m$ are radiating in phase with

$\lambda=0.50m$ . A detector moves in a circular path around the two sources in a plane containing them. How many maxima are detected?

0%
$16$
0%
$20$
0%
$24$
0%
$32$

Explanation

For maxima

$d\sin { \theta } =n\lambda$

On circle, for

$-\dfrac { \pi }{ 2 } \le \theta \le \dfrac { \pi }{ 2 } \Rightarrow n=\dfrac { 2d }{ \lambda } +1$

for

$\dfrac { \pi }{ 2 } <\theta <-\dfrac { \pi }{ 2 } \Rightarrow n=\dfrac { 2d }{ \lambda } -1$

Total number of maxima

$n=\dfrac { 4d }{ \lambda } =\dfrac { 4\left( 2 \right) }{ 0.5 } =16$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

For constructive interference, path difference between interfering beams=

$n\lambda$ .

Here for constructive interference at point P,

$|S_{1}P-S_{2}|=n\lambda$

At point Q,

$|S_{1}Q-S_{2}Q=|S_{1}P-S_{2}P|=n\lambda$

Hence waves interfere constructively at Q.

A radio transmitting station operating at a frequency of

$120$ MHz has two identical antennas that radiate in phase. Antenna

$B$ is

$9m$ to the right of antenna

$A$ . Consider point

$P$ at a horizontal distance

$x$ to the right of antenna

$A$ as shown figure. The value of

$x$ and order for which the constructive interference will occur at point

$P$ are

0%
$x=14.95m,n=1$
0%
$x=5.6m,n=2$
0%
$x=1.65m,n=3$
0%
$x=0,n=3.6$

Explanation

For constructive interference at point P, the path difference between the radio signals at point P=

$n\lambda$

$\implies\sqrt{x^2+81^2}-x=n\lambda$ ,....... Equation 1

$\lambda=\dfrac{c}{\nu}=\dfrac{3\times 10^8}{120\times 10^6}=2.5m$

Using equation 1,

For n=1,

$x=14.95m$

For n=2,

$x=5.6m$

For n=3,

$x=1.65m$

n can have integer values only. Hence option D is not possible.

If a maxima is formed at a detector, then the magnitude of wavelength

$\lambda$ of the wave produced is given by

0%
$\pi R$
0%
$\cfrac{\pi R}{2}$
0%
$\cfrac{\pi R}{4}$
0%
all of these

Explanation

Net path difference between the waves reaching at D,

$x= \dfrac{3}{2}\pi R- \dfrac{\pi R}{2}= \pi R$

For maxima,

$\pi R= n \lambda$

$\lambda= \dfrac{\pi R}{n}$

Thus

$\lambda= \pi R, \dfrac{\pi R}{2}, \dfrac{\pi R}{4}$ and so on.

A parallel beam of light

$(\lambda=500nm)$ is incident at an angle

$\alpha={30}^{o}$ with the normal to the slit plane in Young's double-slit experiment. Assume that the intensity due to each slit at any point on the scree is

${I}_{0}$ . Point

$O$ is equidistant from

${S}_{1}$ and

${S}_{2}$ . The distance between slits is

$1mm$ . Then

0%
the intensity at $O$ is ${I}_{0}$
0%
the intensity at $O$ is zero
0%
the intensity at a point on the screen $1m$ below $O$ is ${I}_{0}$
0%
the intensity at a point on the screen $1m$ below $O$ is zero

Explanation

$Intensity\ at\ O\\ d\sin { \alpha =n\lambda } \\ { 10 }^{ -3 }\sin { 30=500n\times { 10 }^{ -9 } } \\ n=1000\\ \Rightarrow Constructive\ interference\ occurs\\ \\ 1m\ below\ O\\ -d\sin { \alpha +dsin\theta =n\lambda } \\ \tan { \theta =\dfrac { 1 }{ \sqrt { 3 } } } \\ n=0\\ \Rightarrow Constructive\ interference\ occurs\\ \\ \\ \\$

In a Young's double slit experiment set up, source

$S$ of wavelength

$500 nm$ illuminates two slits

$S_{1}$ and

$S_{2}$ which act as two coherent sources. The source

$S$ oscillates about its own position according to the equation

$y = 0.5 \sin \pi t$ where

$y$ is in mm and

$t$ in seconds. The minimum value of time

$t$ for which the intensity at point

$P$ on the screen exactly infront of the upper slit becomes minimum is :

0%
$1 s$
0%
$2 s$
0%
$3 s$
0%
$1.5 s$

Explanation

By SHM, Max speed will be at mean position.

When source is moving towards Slit setup, Number of Photons Hitting the Point P will greater than the situation when the source is moving away from the setup.

Understanding by Droppler's Effect where the velocity of source changes the wavelength of sound.

Intensity will be lowest when the source is at mean position moving away from the setup.

For that

$y = 0 = 0.5 \times \sin (\pi\times t)$

t = Whole Numbers

Velocity of Source:

$\dfrac{\mathrm{d} y}{\mathrm{d} t} = 0.5 \times \pi \times \cos(\pi\times t)$

For Max Velocity:

$\dfrac{\partial^2 { y}}{\partial t^2} = -0.5 \times \pi^{2} \times \sin(\pi\times t) = 0$

t = Whole Numbers

But for t = 1 , Velocity is max and in opposite direction.

Hence t = 1 is the answer

If a minima is formed at the detector, then the magnitude of wavelength

$\lambda$ of the wave produced is given by

0%
$2\pi R$
0%
$\cfrac{3}{2}\pi R$
0%
$\cfrac{5}{2}\pi R$
0%
none of these

Explanation

Path difference at D,

$\Delta x= \pi R$

For minima,

$\pi R= (n+\dfrac{1}{2}) \lambda$

$\implies \lambda= \dfrac{\pi R}{(n+\dfrac{1}{2})}$

Thus,

$\lambda$ can be

$2\pi R, \dfrac{2}{3}\pi R, \dfrac{2}{5}\pi R..........so on$

$z=\cfrac{\lambda D}{4d}$

0%
${ \left[ 3-2\sqrt { 2 } \right] }^{ 2 }$
0%
${ \left[ 3+\sqrt { 2 } \right] }^{ 2 }$
0%
${ \left[ 3-\sqrt { 2 } \right] }^{ 2 }$
0%
${ \left[ 3+2\sqrt { 2 } \right] }^{ 2 }$

Explanation

$S _{3} = 4 I _{o}$

$S _{4} = 2 I _{o}$

$I _{max} = 4I _{o} + 2I _{o} + 2\sqrt{4I _{o}2I _{o}} = (6 + 4\sqrt{2})I _{o}$

$I _{min} = 4I _{o} + 2I _{o} - 2\sqrt{4I _{o}2I _{o}} = (6 - 4\sqrt{2})I _{o}$

Hence

$\dfrac{I _{max}}{I _{min}} = [3 + 2\sqrt{2}]^2$

Consider the optical system shown in the figure that follows. The point source of light

$S$ is having wavelength equal to

$\lambda$ . The light is reaching screen only after reflection. For point

$P$ to be

$2^{nd}$ maxima, the value of

$\lambda$ would be

$(D > > d$ and

$d > > \lambda)$ :

0%
$\dfrac {12d^{2}}{D}$
0%
$\dfrac {6d^{2}}{D}$
0%
$\dfrac {3d^{2}}{D}$
0%
$\dfrac {24d^{2}}{D}$

Explanation

Two rays from light source S get reflected from the two mirrors at M and N respectively and then interfere at point P.

Length of path

$SMP= \sqrt{d^2 + (\dfrac{D}{2})^2} +\dfrac{D}{2}$

Length of path

$SNP= \sqrt{(3d)^2 + (\dfrac{D}{2})^2} +\sqrt{(\dfrac{D}{2})^2+(4d)^2}$

For second maxima to be obtained at P,the path difference must be

$2\lambda$

i-e

$SNP-SMP=$

$[\sqrt{(3d)^2 + (\dfrac{D}{2})^2} +\sqrt{(\dfrac{D}{2})^2+(4d)^2}]-$

$[ \sqrt{d^2 + (\dfrac{D}{2})^2} +\dfrac{D}{2}] = 2\lambda$

Given d<<D

using Bionomial Expansion,

$\sqrt{1+x}=1+\dfrac{x}{2} for \,x<<1$

$\lambda=\dfrac{12d^2}{D}$

The YDSE apparatus is as shown in the figure. The condition for point

$P$ to be a dark fringe is (

$l=$ wavelength of light waves)

0%
$\left( { l }_{ 1 }-{ l }_{ 3 } \right) +\left( { l }_{ 2 }-{ l }_{ 4 } \right) =n\lambda$
0%
$\left( { l }_{ 1 }-{ l }_{ 2 } \right) +\left( { l }_{ 3 }-{ l }_{ 4 } \right) =\cfrac { \left( 2n+1 \right) \lambda }{ 2 }$
0%
$\left( { l }_{ 1 }+{ l }_{ 3 } \right) +\left( { l }_{ 2 }+{ l }_{ 4 } \right) =\cfrac { \left( 2n-1 \right) \lambda }{ 2 }$
0%
$\left( { l }_{ 1 }-{ l }_{ 2 } \right) +\left( { l }_{ 4 }-{ l }_{ 3 } \right) =\cfrac { \left( 2n-1 \right) \lambda }{ 2 }$

Explanation

Path difference between

$SS_1P$ and

$SS_2P$ is

$\Delta = (l_1 + l_3) -( l_2 + l_4) = ( n + \dfrac{1}{2})\lambda$

$\Rightarrow \Delta = (l_1 - l_2) + (l_3 -l_4) = ( n + \dfrac{1}{2})\lambda$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect Reason is correct

Explanation

Resultant intensity at a point due to superposition of two waves of intensity

$I_{1},I_{2}$ at the point is given by,

$I=I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}cos\delta$

For minima,

$\delta=\pi$

Hence,

$I=(\sqrt{I_{1}}-\sqrt{I_{2}})^2$

But at the point of minima, the path travelled by two waves would be different. So they would have difference amplitudes at that point, and so the intensities.

Resultantly the intensity of the superposition would be finite.

A long horizontal slit is place

$1\ mm$ above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen

$1\ m$ away from the slit. If the mirror reflects only

$64\%$ of the light falling on it, the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen is :

0%
$8 : 1$
0%
$3 : 1$
0%
$81 : 1$
0%
$9 : 1$

Explanation

Before reflection, intensity of light is

${I}_{o}$ (say) and after reflection it becomes

${0.64I}_{o}$

${ I }_{ max }={ I }_{ o }+{ 0.64I }_{ o }+2\sqrt { { I }_{ o }\times { 0.64I }_{ o } } =3.24{ I }_{ o }$

${ I }_{ min }={ I }_{ o }+{ 0.64I }_{ o }-2\sqrt { { I }_{ o }\times { 0.64I }_{ o } } =0.04{I}_{o}$

$\frac { { I }_{ max } }{ { I }_{ min } } =\dfrac { 324 }{ 4 } =81:1$

$z=\cfrac{\lambda D}{d}$

0%
$4$
0%
$2$
0%
$\infty$
0%
$1$

Explanation

Intensity at:

$S _{3} = 4 I _{o}$

$S _{4} = 4 I _{o}$

$I _{max} = 16I _{o}$

$I _{min} = 0$

Hence

$\dfrac{I _{max}}{I _{min}} = \infty$

$t=0$ , fringe width is

${\beta}_{1}$ , and at

$t=2s$ , fringe width of figure is

${\beta}_{2}$ . Then

0%
${\beta}_{1}> {\beta}_{2}$
0%
${\beta}_{2}> {\beta}_{1}$
0%
${\beta}_{1}= {\beta}_{2}$
0%
data is insufficient

Explanation

The position of source at

$t=0s$ is:

$y(0)=1+cos0=2mm$

The position of source at

$t=2s$ is:

$y(0)=1+cos2\pi=2mm$

Hence, at the two different times, the position of the source is the same, and hence interference pattern formed is same and so is the fringe width.

Hence,

$\beta_1=\beta_2$

$t=2s$ , the position of central maxima is

0%
$2mm$ above $C$
0%
$2mm$ below $C$
0%
$4mm$ above $C$
0%
$4mm$ below $C$

Explanation

Path Difference from Source:

$\delta _{Source} = \dfrac{1\times 10^{-3}\times 2\times 10^{-3}}{1}$

Path Difference on the screen:

$\delta = \dfrac{x\times 2\times 10^{-3}}{2}$

For Central Maxima:

$\delta _{Source} = \delta$

Which gives

$x = 2mm$

And as the source is above, the central fringe will be below the central horizontal axis.

Answer is B: 2mm below C

In the arrangement shown in figure,

$D>>d$ . For what minimum value of

$d$ is there a dark band at point

$O$ on the screen?

0%
$\sqrt{\cfrac{D\lambda}{4}}$
0%
$\sqrt{\cfrac{3D\lambda}{4}}$
0%
$\sqrt{\cfrac{D\lambda}{8}}$
0%
$\sqrt{\cfrac{2D\lambda}{3}}$

Explanation

Path difference between A and B is given by

$\Delta x =\dfrac{dy}{D}$ where

$y=d$

net path difference

$= 2 \Delta x =\dfrac{2d^2}{D}$

For minima at O it should be multiple of

$\dfrac{\lambda}{2}$

$\dfrac{2d^2}{D}=\dfrac{\lambda}{2}$

$d=\sqrt{\dfrac{\lambda D}{4}}$

A liquid of refractive index

$\mu$ is filled between the screen and slits.

0%
$\cfrac { 2\pi }{ \lambda } \left[ \left[ \sqrt { { d }^{ 2 }+{ x }_{ 0 }^{ 2 } } +{ x }_{ 0 } \right] +\cfrac { \mu { d }^{ 2 } }{ 2D } \right]$
0%
$\cfrac { 2\pi }{ \lambda } \left[ \left[ \sqrt { { d }^{ 2 }+{ x }_{ 0 }^{ 2 } } -{ x }_{ 0 } \right] +\cfrac { \mu { d }^{ 2 } }{ 2D } \right]$
0%
$\cfrac { 2\pi }{ \lambda } \left[ \left[ \sqrt { { d }^{ 2 }-{ x }_{ 0 }^{ 2 } } +{ x }_{ 0 } \right] +\cfrac { \mu { d }^{ 2 } }{ 2D } \right]$
0%
$\cfrac { 2\pi }{ \lambda } \left[ \left[ \sqrt { { d }^{ 2 }-{ x }_{ 0 }^{ 2 } } -{ x }_{ 0 } \right] +\cfrac { \mu { d }^{ 2 } }{ 2D } \right]$

Explanation

Path Difference due to Position of source:

$\delta _{source} = \sqrt{d^{2}+x_{o}^{2}} - x _{o}$

Path Difference at O is Zero

Path Difference at P:

$\delta _{P} = \dfrac{d}{2}\times\dfrac{d}{D}$

For Phase Difference: Multiply

$\delta _{source}$ by

$\dfrac{2\pi}{\lambda}$ and

$\delta _{P}$ by

$\dfrac{2\pi\mu}{\lambda}$

Therefore, Answer is

$\phi = \dfrac{2\pi}{\lambda}(\delta _{source} + \mu\times\delta _{P})$

$\Rightarrow \phi =\cfrac { 2\pi }{ \lambda } \left[ \left[ \sqrt { { d }^{ 2 }+{ x }_{ 0 }^{ 2 } } -{ x }_{ 0 } \right] +\cfrac { \mu { d }^{ 2 } }{ 2D } \right]$

$z=\cfrac{\lambda D}{2d}$

0%
$1$
0%
$1/2$
0%
$3/2$
0%
$2$

Explanation

Intensity at:

$S _{3} = 4 I _{o}$

$S _{4} = 0$

$I _{max} = 4I _{o}$

$I _{min} = 4I _{o}$

$\therefore \dfrac{I _{max}}{I _{min}} = 1$

If the incident beam makes an angle of

${30}^{o}$ with the x-axis (as in the dotted arrow shown in the figure), find the y-coordinates of the first minima on either side of the central maximum..

0%
$\cfrac { 3 }{ \sqrt { 7 } }$ and $\cfrac { 1 }{ \sqrt { 15 } }m$
0%
$\cfrac { 3 }{ \sqrt { 7 } }$ and $\cfrac { 2 }{ \sqrt { 15 } }m$
0%
$\cfrac { 3 }{2 \sqrt { 7 } }$ and $\cfrac { 1 }{ \sqrt { 15 } }m$
0%
$\cfrac { 6 }{ \sqrt { 7 } }$ and $\cfrac { 3 }{ \sqrt { 15 } }m$

Explanation

Path Difference due to Angle:

$\delta _{source} = 0.5mm$

For Central Maxima, Path Difference should be Zero:

$\delta _{source} = x\times\dfrac{d}{D}$

which gives us

$x = 0.5m$

In the arrangement shown in figure,

$D>>d$ . Find the fringe width.

0%
$d$
0%
$2d$
0%
$4d$
0%
$3d$

In the arrangement shown in figure,

$D>>d$ . Find the distance

$x$ at which the next bright fringe is formed.

0%
$\cfrac{3\lambda}{2}$
0%
$\cfrac{\lambda}{4}$
0%
$\cfrac{\lambda}{2}$
0%
$\cfrac{5\lambda}{2}$

Determine the width of the region where the fringes will be visible

0%
$4cm$
0%
$6cm$
0%
$2cm$
0%
$3cm$

Explanation

Rays emerging from point source S get reflected on the mirror DC at points D and C ,and then reach point L and N respectively on the screen.

LN is the region where the fringes will be visible.

In the similar triangle SDP and LDO,

$\dfrac{y_2}{1}=\dfrac{195}{5}$

$y_2= 39 mm$

In similar triangle SCP and NCO,

$\dfrac{y_1}{1} =\dfrac{190}{10}$

$y_1= 19mm$

$y_2-y_1= 20 mm$

Width of the region where the fringes will be visible is

$2 cm$

Find the fringe width of the fringe pattern.

0%
$0.05cm$
0%
$0.25cm$
0%
$0.01cm$
0%
$0.1cm$

Explanation

$S'$ is the virtual point source such that light from

$S$ and

$S'$ interfere on the the screen, then this situation resembles with the Young's double slit experiment.

Let d be the distance between the two point source.

$d=2 SP= 2 mm$

$D=2m v=3 \times 10^8 ms^{-1} \nu = 6\times 10^{14} Hz$

using

$v = \nu \lambda$

On solving,

$\lambda = 500 nm$

fringe width,

$\beta= \dfrac{\lambda D}{d}$

This gives

$\beta = .05 cm$

The fractional change in intensity of the central maximum as function of time is

0%
$\cfrac { A\sin { \omega t } }{ L }$
0%
$\cfrac { 2A\sin { \omega t } }{ L }$
0%
$\cfrac { 3A\sin { \omega t } }{ L }$
0%
$\cfrac { 4A\sin { \omega t } }{ L }$

Explanation

$I=2{ I }^{ 1 }(1+\cos { \delta } )$

Where

$\delta =phase$

$\delta =\mu \dfrac { w }{ c } \triangle x$

But,

$\triangle x=A \sin { wt }$

$\therefore \delta$ is small thus,

$I=\dfrac { 2A \sin { wt } }{ L }$

When the source comes toward the point

$Q$ ,

0%
the bright fringes will be less bright
0%
the dark fringes will no longer remain dark
0%
the fringe width will increase
0%
none of these

Light of wavelength

$6328 \overset{o}{A}$ is incident normally on a slit having a width of

$0.2 mm$ . The angular width of the central maximum measured from minimum to minimum of diffraction pattern on a screen

$9.0$ meters away will be about

0%
$0.36 degree$
0%
$0.18 degree$
0%
$0.72 degree$
0%
$0.09 degree$

Explanation

The angular width of central maxi. is
2

$\theta$ = 2

$\displaystyle\dfrac{\lambda}{a}$ =

$\displaystyle\dfrac{2 \times 6328 \times 10^{-10}}{2 \times 10^{-4}}$ radian.

$= 6328 \times$ 10

$^{-6} \times \displaystyle\dfrac{180}{\pi}$

$degree = 0.36^{o}$

Calculate the number of fringes.

0%
$10$
0%
$20$
0%
$30$
0%
$40$

Explanation

As calculated above,

$\beta= .05 cm$

Region for which fringes are visible,

$y = 2 cm$

Number of fringes,

$N = \dfrac{y}{\beta}$

This gives

$N = 40$

The position of the direct image obtained at O

$_3$ , when a monochromatic beam of light is passed through a plane transmission grating at normal incidence as shown in Fig. The diffracted images A, B and c correspond to the first, second and third order diffraction. when the source is replaced by another source of shorter wave-length

0%
all the four will shift in the direction C to O
0%
all the four will shift in the direction O to C
0%
the images C, B and A will shift towards O
0%
the images C,B and A will shift away from O

Explanation

For a diffraction grating, with distance 'd' between adjacent slits, and

$\theta_{n}$ the angle between normal to the grating and the diffracted wavelength for nth order,

$d\sin \theta_{n}=n\lambda$

$\lambda$ is reduced,

$\sin\theta_{n}$ becomes smaller. Thus the angle of diffraction for all order reduces, hence the pattern shrinks and images move towards the normal, not diffracted beam.

Hence the images shift towards O.

A point source is emitting light of wavelength 6000

$\overset{o}{A}$ is placed at a very small height h above a flat reflecting surface

$MN$ as shown in the figure. The intensity of the reflected light is

$36$ % of the incident intensity. Inference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance

$D$ from it. If the intensity at

$p$ be maximum, then the minimum distance through which the reflecting surface

$MN$ should be displaced so that at

$P$ again becomes maximum?

0%
3 $\times$ 10 $^{-7}$ m
0%
6 $\times$ 10 $^{-7}$ m
0%
1.5 $\times$ 10 $^{-7}$ m
0%
12 $\times$ 10 $^{-7}$ m

Explanation

Light reaches to point P by two ways.

First, directly from source S and second, after reflection from mirror MN. Therefore path difference,

$\Delta x=2h$ ,

Now, let the surface MN is displaced by

$y$ .

Hence the path difference,

$\Delta x'=2(h\pm y)$ ,

Extra path difference due to displacement of surface MN,

$\Delta x''=2(h\pm y)-2h=\pm2 y$

The intensity of light at P will be maximum again, if

$2y=\lambda=6000$ ,

$y=6000/2=3000A^{o}$ ,

$y=3\times10^{-7}m$

Two identical coherent sources are placed on a diameter of a circle of radius R at a separation d (d << R, d >>

$\lambda$ ) symmetrically about the centre of the circle. The sources emit identical wavelength

$\lambda$ . The number of the points on the circle with maximum intensity is

0%
$\displaystyle \frac{2d}{\lambda }+1$
0%
$\displaystyle \frac{4d}{\lambda }$
0%
$\displaystyle \frac{4d}{\lambda }-2$
0%
$\displaystyle \frac{4d}{\lambda }+2$

Explanation

For maxima

$d\sin { \theta } =n\lambda$

On circle, for

$-\dfrac { \pi }{ 2 } \le \theta \le \dfrac { \pi }{ 2 } \Rightarrow n=\dfrac { 2d }{ \lambda } +1$

for

$\dfrac { \pi }{ 2 } <\theta <-\dfrac { \pi }{ 2 } \Rightarrow n=\dfrac { 2d }{ \lambda } -1$

Total number of maxima

$n=\dfrac { 4d }{ \lambda }$

In Young's double slit experiment if the slit widths are in the ratio

$1:9$ . The ratio of the intensity at minima to that at maxima will be :

0%
$1$
0%
$\dfrac{1}{9}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{3}$

Ratio of maximum to minimum intensities at

$P$ is

0%
$2 : 1$
0%
$4 :1$
0%
$8 : 1$
0%
$16 : 1$

Explanation

Before reflection, intensity of light is

${I}_{o}$ (say) and after reflection it becomes

${0.36I}_{o}$

${ I }_{ max }={ I }_{ o }+{ 0.36I }_{ o }+2\sqrt { { I }_{ o }{ 0.36I }_{ o } } =2.56{ I }_{ o }$

${ I }_{ min }={ I }_{ o }+{ 0.36I }_{ o }-2\sqrt { { I }_{ o }{ 0.36I }_{ o } } =0.16{I}_{o}$

$\dfrac { { I }_{ max } }{ { I }_{ min } } =\dfrac { 256 }{ 16 } =16:1$

An interference is observed due to two coherent sources 'A' & 'B' separated by a distance of 4

$\lambda$ along the y-axis where

$\lambda$ is the wavelength of the source. A detector D is moved on the positive x-axis. The number of points on the x-axis excluding the points, x =0 & x =

$\infty$ at which maximum will be observed is

0%
three
0%
four
0%
two
0%
infinite

Explanation

From

$\triangle ABD$

$(AD)^2=(AB)^2+(BD)^2$

$\implies (AP+PD)^2=(AB)^2+(BD)^2$

Here

$AP$ is the path difference.

$\therefore AP=BD=x_n$

$\therefore (n\lambda+x_n)^2=(4\lambda)^2+(x_n)^2$

$x_n=\dfrac{16\lambda^2-n^2\lambda^2}{2n\lambda}$

$\therefore x_1=\dfrac{16\lambda^2-\lambda^2}{2\lambda}=7.5\lambda$

$x_2=\dfrac{16\lambda^2-4\lambda^2}{4\lambda}=3\lambda$

$x_3=\dfrac{16\lambda^2-9\lambda^2}{6\lambda}=\dfrac76\lambda$

$x_4=0$

No of points where maxima is observed=

$3$

The shape of the interference fringes, on the screen is

0%
circle
0%
ellipse
0%
parabola
0%
straight line

Explanation

$S'$ represents the another point source forms due to reflection from the mirror.

$S'P= \sqrt{(x+h)^2+ y^2}$ and

$SP= \sqrt{(x-h)^2+ y^2}$
Let path difference at point P be

$\Delta$
Using,

$\Delta + SP= S'P$
Squaring and adding,

$\Delta^2 + 4hx= -2\Delta\sqrt{(x-h)^2+ y^2}$
Squaring and adding again,

$\Delta^4 + 4h^2x^2 + 8hx\Delta= 4\Delta^2(x^2 +h^2- 2hx + y^2)$

$\implies 4(\Delta^2-h^2)x^2 - (8\Delta^2h+8h\Delta)x + y^2+ 4\Delta^2h^2-\Delta^4= 0$
which represents equation of circle as

$X^2 + Y^2= R^2$
Thus the fringes will of Circular shape centered at O.

Light is incident at an angle

$\displaystyle \phi$ with the normal to a plane containing two slits of separation d Select the expression that correctly describe the positions of the interference maxima in terms of the incoming angle

$\displaystyle \theta$ and outgoing

$\displaystyle \phi$

0%
$\displaystyle \sin \phi \sin \theta =\left ( m+\frac{1}{2} \right )\frac{\lambda }{d}$
0%
$\displaystyle d\sin \theta =m\lambda$
0%
$\displaystyle \sin \phi -\sin \theta (m+1)\frac{\lambda }{d}$
0%
$\displaystyle \sin \phi +\sin \theta =m\frac{\lambda }{d}$

In Young's double slit experiment, if the widths of the slit are in the ratio

$4:9$ , ratio of intensity of maxima to intensity of minima will be

0%
$25:1$
0%
$9:4$
0%
$3:2$
0%
$81:16$

Explanation

$$\textbf{HINT:}$$ the measurable amount of a property, such as force, brightness, or a magnetic field.

$\textbf{Step1:}$ width of silt is directly proptiontal to intensity of light

Width of slit $(\mathrm{W}) \propto$ Light intensity (I), $\dfrac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\dfrac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\dfrac{4}{9}=$ $\dfrac{4 \mathrm{k}}{9 \mathrm{k}} \quad \ldots(\mathrm{k}$ is a constant $)$

$I_{\max }=4 \mathrm{k}+9 \mathrm{k}+2 \sqrt{4 \mathrm{k} 9 \mathrm{k}}=25 \mathrm{k} \\$

$\mathrm{I}_{\min }=4 \mathrm{k}+9 \mathrm{k}-2 \sqrt{4 \mathrm{k} 9 \mathrm{k}}=\mathrm{k} \\$

$\textbf{Step 2:}$ Dividing the expression,

$\dfrac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=25:1$

A plane wave of monochromatic light falls normally on a uniformly thin film of oil which covers a glass plate. The wavelength of source constructive interference is observed for

$\lambda_{1} = 5000\overset {\circ}{A}$ and

$\lambda_{2} = 10000\overset {\circ}{A}$ and for no other wavelength in between. If

$\mu$ of oil is

$1.25$ and that of glass is

$1.5$ , the thickness of film will be ________

$\mu m$ .

0%
$0.2$
0%
$0.1$
0%
$0.8$
0%
$0.4$

Explanation

Given:

$\mu_o=1.25, \mu_{g}=1.5, \lambda_1 = 5 \times 10^{-7}m=\lambda_{n+1}$ (No wavelength between destructive interference)

$\lambda_2=10\times10^{-7}m$

The set u is similar to anti-reflective coating.

For destructive interference

$2\mu \cos r=(2n-1)\dfrac {\lambda_n}2=[2(n+1)-1]\dfrac {\lambda_{n+1}}2......(i)$

$(2n-1)\dfrac {5\times 10^{-7}}{2}=(2n+1)\dfrac {10\times10^{-7}}{2}$

$n=\dfrac 32$ ....(ii)

For, normal incidence,

$cos r =1,$

from (i) and (ii)

$2\mu t=\left(2\times \dfrac 32-1\right)\dfrac {\lambda_n}{2}=\lambda_n=5\times 10^{-7}$

$t=\dfrac {5\times 10^{-7}}{2\times1.25}=2\times 10^{-7}m$

or the thickness =

$0.2\mu m$

CBSE Questions for Class 12 Medical Physics Wave Optics Quiz 13 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Wave Optics Quiz 13 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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