MCQ Questions for Class 12 Medical Physics Wave Optics Quiz 9

Two coherent monochromatic light beams of intensities

$I$ and

$4I$ are superposed. The maximum and minimum possible intensities in the resulting beam are:

0%
$5I$ and $I$
0%
$5I$ and $3I$
0%
$9I$ and $I$
0%
$9I$ and $3I$

Explanation

The correct answer is option (C).

Hint: Apply the formula of maximum and minimum intensities.

Step 1: Calculate the maximum intensity.

$I_{max}=(\sqrt {I_1}+\sqrt {I_2})^2$

$I_{max}=(\sqrt{I}+\sqrt{4I})^2$

$I_{max}=(\sqrt{I}+2\sqrt{I})^2$

$I_{max}=(3\sqrt I)^2$

$I_{max}=9I$

Step 2: Calculate the minimum intensity.

$I_{min}=(\sqrt{I_1}-\sqrt{I_2})^2$

$I_{min}=(\sqrt{I}-\sqrt{4I})^2$

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

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

$I_{min}=I$

Hence, the maximum and minimum intensities are: $9I$ and $I$ respectively.

The correct answer is option (C).

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

$2m$ away. The distance between the first dark fringes on either side of the central bright fringe is

0%
$1.2\ cm$
0%
$1.2\ mm$
0%
$2.4\ cm$
0%
$2.4\ mm$

Explanation

$\textbf{Given}:$

$\lambda = 600nm$ , d = 1.00 mm, D = 2 m

$\textbf{Solution}:$

For the diffraction at a single slit, the position of minima is given by

$d sin \theta = n \lambda$

For small value of

$\theta$ ,

$sin \theta\approx \theta=\dfrac{y}{D}$

$\therefore \dfrac{dy}{d}=\lambda$ or

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

Substituting the values ,we have

$y=\dfrac{2 \times 6 \times 10^{-7}}{1 \times 10^{-3}}=1.2 \times 10^{-3}m=1.2 mm$

$\therefore$ Distance between first minima on either side of central side of central maxima=2y=2.4mm.

$\textbf{Hence D is the correct option}$

n identical waves each of intensity

${ 1 }_{ 0 }$ interfere with each other. The ratio of maximum intensities if the interference is (i) coherent and (ii) incoherent is:

0%
${ n }^{ 2 }$
0%
$\frac { 1 }{ n }$
0%
$\frac { 1 }{ { n }^{ 2 } }$
0%
n

Explanation

In case of coherent source

${ l }_{ max }={ \left( \sqrt { { l }_{ 0 } } +\sqrt { { l }_{ 0 } } +\sqrt { { l }_{ 0 } } +.....upto\quad n\quad times \right) }^{ 2 }={ n }^{ 2 }{ l }_{ 0 }$

In case of incoherent source,

${ l }_{ max }=n{ l }_{ o }$

$\therefore$ Ratio=n

Angular - width of central maximum in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength

$6000\overset {\circ}{A}$ . When the slit is illuminated by light of another wavelength the angular-width decreases by

$30\%$ . Calculate the wavelength of this light. The same decrease in the angular-width of central maximum is obtained when the original apparatus is immersed in a liquid. Find refractive index of the liquid.

0%
$4200, 21.43$ .
0%
$5200, 2.43$ .
0%
$14200, 1.43$ .
0%
$4200, 1.43$ .

Explanation

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

i.e.,

$\beta \propto \lambda$

So,

$\cfrac { 6000 }{ \lambda } =\cfrac { 100 }{ 70 } \\ \lambda =4200\mathring { A }$

And

$\mu =\cfrac { { \lambda }_{ vaccum } }{ { \lambda }_{ medium } } \\ \mu =\cfrac { 6000 }{ 4200 } \\ =1.43$

$I_0$ is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled?

0%
$2I_0$
0%
$4I_0$
0%
$I_0$
0%
$I_0 / 2$

Explanation

When the slit width is doubled the energy falling per sec on the screen is doubled but angular width of central maxima

$\left( \Delta \theta = \dfrac{2 \lambda}{b} \right)$ is halved. Hence intensity of the central maxima will be

$4I_0$ .

The resultant amplitude of a vibrating particle by the superposition of the two waves

$y_{1}= a \sin\left(\omega t+\dfrac{\pi}{3}\right)$ and

$y_{2}=a \sin \omega t$ is :

0%
$a$
0%
$\sqrt{2}a$
0%
$2a$
0%
$\sqrt{3}a$

Explanation

${ y }_{ 1 }=a\sin { \left( \omega t+\cfrac { \pi }{ 3 } \right) } \quad { y }_{ 2 }=a\sin { \omega t } \quad\quad\theta =\omega t+\cfrac { \pi }{ 3 } -\omega t\\ \theta =\cfrac { \pi }{ 3 }$

Resultant amplitude

$=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+2{ a }_{ 1 }{ a }_{ 2 }\cos { \theta } } \\ =\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+2{ a }_{ 1 }{ a }_{ 2 }\cos { \cfrac { \pi }{ 3 } } } \\ =\sqrt { 3 } a$

A transparent paper ( refractive index

$=1.45$ ) of thickness

$0.02$ mm is pasted on one of the slits of a young's double slit experiment which uses monochromatic light of wavelength

$620$ nm. How many fringes will pass through the center if the paper is removed ?

0%
14
0%
15
0%
18
0%
44

Explanation

given,

$\mu =1.45$

$t=0.02\times10^{-3}m$

$\lambda =620\times10^{-9}m$

$n=\dfrac{(\mu-1)t}{\lambda}$

$\Rightarrow =\dfrac{(1.45-1)\times0.02\times10^{-3}}{620\times10^{-9}}$

$n=\dfrac{0.45\times0.02\times10^{-3}}{620\times10^{-9}}$

$=14.25$

A photograph of the moon was taken with telescope. Later on, it was found that a housefly was siting on the objective lens of the telescope. In photograph

0%
the image of the housefly will be reduced
0%
there is a reduction in the intensity of the image
0%
there is an increase in the intensity of the image
0%
the image of the housefly will be enlarged

A light of wavelength 6300A shine on a two narrow slits separated by a distance 1.0 mm and illuminates a screen at a distance distance 1.5 m away. When one slit is covered by a thin glass of refractive index 1.8 and other slit by a thin glass plate of refractive index

$\Pi$ , the central maxima shifts by

${ 6 }^{ \circ }$ . Both plates have same thickness of 0.5 mm. The value of refractive index

$\Pi$ of the plate is

0%
1.6
0%
1.7
0%
1.5
0%
1.4

Ratio of intensities of two waves are given by

$4:1$ . Than the ratio of the amplitude of the two waves is:

0%
$2:1$
0%
$1:2$
0%
$4:1$
0%
$1:4$

Two coherent sources of wavelength

$6.2\times 10^{-7}$ m produce interference. The path difference corresponding to

$10^{th}$ order maximum will be?

0%
$6.2\times 10^{-6}$ m
0%
$3.1\times 10^{-6}$ m
0%
$1.5\times 10^{-6}$ m
0%
$12.4\times 10^{-6}$ m

Explanation

for

${ m^{ th } }$ order maximum path difference

$=n\lambda =10\times 6.2\times { 10^{ -7 } }\, m \\ =6.2\times { 10^{ -6 } } m$

Option is A

A screen is at a distance of 2 m from a narrow slit illuminated with light of 600 nm. The first minimum lies 5 mm on either side of the central maximum. The width of slit is.

0%
24 mm
0%
0.24 mm
0%
2.4 mm
0%
0.024 mm

The interference pattern with two coherent light sources of density ratio

$n$ . In the interference pattern, the ratio

$\dfrac {I_{max}-I_{min}}{ I_{max}+I_{min}}$ will be:

0%
$\dfrac {\sqrt {n}}{n+1}$
0%
$\dfrac {2 \sqrt {n}}{n+1}$
0%
$\dfrac {\sqrt {n}}{(n+1)^{2}}$
0%
$\dfrac {2 \sqrt {n}}{(n+1)^{2}}$

Explanation

Given that

$\cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =n\\ { I }_{ max }={ I }_{ 1 }+{ I }_{ 2 }+2\sqrt { { I }_{ 1 }{ I }_{ 2 } } \\ { I }_{ min }={ I }_{ 1 }+{ I }_{ 2 }-2\sqrt { { I }_{ 1 }{ I }_{ 2 } } \\ \therefore \cfrac { { I }_{ max }+{ I }_{ min } }{ { I }_{ max }-{ I }_{ min } } =\cfrac { 4\sqrt { { I }_{ 1 }{ I }_{ 2 } } }{ 2\sqrt { { I }_{ 1 }{ I }_{ 2 } } } \\ \quad \quad \quad \quad \quad \quad =\cfrac { 4{ I }_{ 2 }\sqrt { { I }_{ 1 }/{ I }_{ 2 } } }{ 2{ I }_{ 2 }\sqrt { { I }_{ 1 }/{ I }_{ 2 }+1 } } \\ \quad \quad \quad \quad \quad \quad =\cfrac { 2\sqrt { n } }{ n+1 } \\$

In a double slit experiment, the separation between the slits is d = 0.25 cm and the distance of the screen D = 100 cm from the slits, If the wavelength of light used is

$\lambda = 6000\mathop {\text{A}}\limits^{\text{o}} \,{\text{and}}\,{{\text{I}}_{\text{0}}}$ is the intensity of the central bright fringe, the intensity at a distance

$y = 4 \times {10^{ - 5}}$ m from the central maximum is

0%
${I_0}$
0%
${I_0}$ /2
0%
3 ${I_0}$ /4
0%
${I_0}$ /3

Explanation

It is given that,

$d=0.25\,cm$

$D=100\,cm$

$\lambda =6000\,\dot{A}=6\times {{10}^{-7}}\,m$

For central maximum

$x=\dfrac{Dn\lambda }{d}$

$x=\dfrac{100\times 6\times {{10}^{-7}}}{0.25}$

$x=4\times {{10}^{-5}}$

Intensity at this point is ${{I}_{0}}$

At $y=4\times {{10}^{-5}}\,m$

Intensity will be ${{I}_{0}}$

$\dfrac{I_{1}}{I_{2}}=\dfrac{9}{1}$ then

$\dfrac{I_{max}}{I_{min}}=?$

0%
$100 : 64$
0%
$64 : 100$
0%
$4 : 1$
0%
$1 : 4$

Explanation

$\cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =\cfrac { 9 }{ 1 } ={ \left( \cfrac { { a }_{ 1 } }{ { a }_{ 2 } } \right) }^{ 2 }\quad \quad \cfrac { { I }_{ max } }{ { I }_{ min } } ={ \left( \cfrac { { a }_{ 1 }+{ { a }_{ 2 } } }{ { a }_{ 1 }-{ a }_{ 2 } } \right) }^{ 2 }\\ \cfrac { { I }_{ max } }{ { I }_{ min } } ={ \left( \cfrac { \cfrac { { a }_{ 1 } }{ { a }_{ 2 } } +{ 1 } }{ \cfrac { { a }_{ 1 } }{ { a }_{ 2 } } -1 } \right) }^{ 2 }={ \left( \cfrac { 3+1 }{ 3-1 } \right) }^{ 2 }={ 2 }^{ 2 }:1\quad =4:1$

A small plane mirror is placed at the center of the spherical screen of radius R. A beam of light is falling on the mirror. If the mirror makes n revolutions per second, the speed of light on the screen after reflection from the mirror will be:

0%
$4\pi nR$
0%
$2\pi nR$
0%
$nR/2\pi$
0%
$nR/4\pi$

For the sustained interference of light, the necessary condition is that the two sources should:

0%
Have constant phase difference
0%
Be narrow
0%
Be close to each other
0%
Of same amplitude

Ray optics is valid when characteristic dimension are

0%
much smaller than wavelength of light.
0%
of same order as wavelength of light
0%
much larger than wavelength of light
0%
none of the above

In a young's double slit experiment , the distance between the two slits is 0.1mm and the wavelength of light used is 4

$\times 10^{-7}$ m . If the width of the fringe on the screen is 4mm, then the distance between screen and slit is

0%
0.1mm
0%
1cm
0%
0.1 cm
0%
1m

The position of image of

$S$ by the mirror is

0%
$2\ mm$ below $S$
0%
$1\ mm$ below
0%
$4\ mm$ below $S$
0%
$None\ of\ these$

In Young's double slit experiment, the ratio of intensities of bright and dark frings is

$9$ . This means that

0%
the intensities of individual source are $5$ and $4$ units respectively
0%
the intensities of individual source are $4$ and $1$ units respectively
0%
the ratio of their amplitude is $3$
0%
the ratio of their amplitude is $4$

Explanation

$\dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{{{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}}{{{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}}=\dfrac{9}{1}$

$\dfrac{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}=\dfrac{3}{1}$

$\sqrt{\dfrac{{{I}_{1}}}{{{I}_{2}}}}=\dfrac{2}{1}$

$\dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{4}{1}$

So, the ratios of their intensities is $4:1$

After reflection from a concave mirror, a plane wavefront becomes

0%
Cylindrical
0%
Spherical
0%
Remains planar
0%
None of the above

To demonstrate the phenomenon of interference we require two sources which emit radiation of

0%
nearly the same frequency
0%
the same frequency
0%
different wavelength
0%
the same frequency and having a definite phase relationship

To observe diffraction. the size of the obstacle

0%
Should be $\lambda$ /2, Where $\lambda$ is the wavelength
0%
Should be Of the order of wavelength
0%
Has no relation to wavelength
0%
Should be much larger than the wavelength

Light of wavelength

$\lambda$ from a point source falls on a small circular obstacle of diameter d. Dark and bright circular rings around a central bright spot are formed on a screen beyond the obstacle. The distance between the screen and obstacle is D. Then , the condition for the formation of rings, is

0%
$\sqrt { \lambda } \approx \cfrac { { d } }{ 4D }$
0%
$\lambda \approx \cfrac { { d }^{ 2 } }{ 4D }$
0%
$d\approx \cfrac { { \lambda }^{ 2 } }{ D }$
0%
$\lambda \approx \cfrac { D }{ 4 }$

A wave or a pulse is reflected normally from the surface of a denser medium back into the rarer medium. The phase change caused by the reflection-

0%
$0$
0%
$\pi /2$
0%
$\pi$
0%
$3\pi /2$

Explanation

When a wave or a pulse is reflected normally from the surface of a denser medium back into the rarer medium, then tere is a phase difference of

$\pi$ . According to electromagnetic theory When a light beam falls on a boundary of two media, then one part of it will be reflected and other part will be refracted. If refractive index of a medium, where the electromagnetic wave is refracted, is greater than the medium incident, then phase of reflected ray will be reverse with respect to the incident beam. As a result of which the wave suffers a phase change of

$\pi$ .

A small coin is resting on the bottom of a beaker filled with a liquid. A ray of light from the coin travels upto the surface of the liquid and moves along its surface (see figure)How fast is the light travelling in the liquid?

0%
$1.2\times10^{8}\ m/s$
0%
$1.8\times10^{8}\ m/s$
0%
$2.4\times10^{8}\ m/s$
0%
$3.0\times10^{8}\ m/s$

Explanation

Hint :-The critical angle is that angle of incidence for which the angle of refraction in rarer medium is equal to 90.

Step 1: Note the given values and consideration

Let angle of incidence = $\theta$ ,

Angle of refraction = $90^o$ ,

Refractive index of liquid $=\mu$ ,

Refractive index of air =1

Step 2: Calculate refractive index of liquid

Using Snell's law

$\mu \sin \theta =1 \times \sin90$

$\mu \dfrac{3}{5}=1$

$\mu =\dfrac{5}{3}$

Step 3: Calculate velocity of light in liquid

$\dfrac{c}{v}=\mu$

$\dfrac{c}{v}=\dfrac{5}{3}$

$v=\dfrac{3\times 3\times {{10}^{8}}}{5}$

$v=1.8\times {{10}^{8}}m/s$

2180407_1077935_ans_4b2473d4909742afbf1e53c4ffe31fc4.png

In Young's double slit experiment, the two slits are

$0.2 mm$ apart. The interference fringes for light of wavelength

$6000 \mathring{A}$ are found on the screen

$80 cm$ away. The distance of fifth dark fringe, from the central fringe, will be:

0%
$6.8 \ mm$
0%
$7.8 \ mm$
0%
$9.8 \ mm$
0%
$10.8 \ mm$

In YDSE The intensity of central bright fringe is

$8mW/m^2$ . What will be the intensity at

$\lambda /6$ path difference?

0%
$8 mW/m^2$
0%
$6 mW/m^2$
0%
$4 mW/m^2$
0%
$2 mW/m^2$

Explanation

Given that,

Intensity of central bright fringe ${{I}_{0}}=8mW/{{m}^{2}}$

We know that,

$I={{I}_{0}}{{\cos }^{2}}\left( \frac{\phi }{2} \right)$

$I=8{{\cos }^{2}}\left( \frac{\phi }{2} \right)$

Now,

Phase difference = $\dfrac{2\pi }{\lambda }$ x path difference

$\phi =\dfrac{2\pi }{\lambda }\times \dfrac{\lambda }{6}$

$\phi =\dfrac{\pi }{3}$

Now, the intensity is

$I=8\times {{\cos }^{2}}\left( \dfrac{\pi }{6} \right)$

$I=8\times \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}$

$I=6\,mW/{{m}^{2}}$

Hence, the intensity is $6\,mW/{{m}^{2}}$

Wavelength of light used in an optical instrument are

$\lambda_1=4000\overset{o}{A}$ and

$\lambda_2=5000\overset{o}{A}$ , then ratio of their respective resolving powers(corresponding to

$\lambda_1$ and

$\lambda_2$ ) is?

0%
$16.25$
0%
$9:1$
0%
$4:5$
0%
$5:4$

Explanation

Resolving Power (R.P) :It is the reciprocal of resolving limit<br>

$R. P. = \dfrac{1} {d\theta}$

For a Telescope : Resolving limit,

$d\theta$

$=$

$\dfrac{1.22 \times \lambda}{a}$

$\therefore R. P. \propto \frac{1} {\lambda}$

Where

$\lambda$ = wavelength of light used to illuminate the object.

$a$ =aperture of the objective.

Given:

$\lambda_1 = 4000 A$ and

$\lambda_2 = 5000A$

$R. P.| \propto \dfrac{1}{\lambda}| \Rightarrow \dfrac{(R.P.)_1}{(R.P.)_2}=\dfrac{\lambda_2}{\lambda_1}=\dfrac 54$

When light waves suffer reflection at the interface between air and glass , the change of phase of the reflected wave is equal to

0%
Zero
0%
$\dfrac{\pi}{2}$
0%
$\pi$
0%
$2\pi$

The sensor is exposed for

$0.1 s$ to a

$200 W$ lamp

$10 m$ away. The sensor has opening that is

$20 mm$ in a diameter. How many photons enter the sensor if the wavelength of light is

$600 mm$ ? (assume that all the energy of the lamp is given off as light).

0%
$1.53 \times 10^{11}$
0%
$1.53 \times 10^{2}$
0%
$1.53 \times 10^{4}$
0%
$1.53 \times 10^{13}$

Explanation

(Refer to Image)

Energy incident on sensor=

$n \cfrac {hc}{x}$

$=\cfrac {\pi\left(\cfrac {20}{1000}\right)\times \cfrac {1}{4}}{4 \pi (10)^2}\times 200 \times \cfrac {1}{10}=x \times \cfrac {6.626}{10^{34}}\times \cfrac {3 \times 100}{600}$

$\Rightarrow n= 1.53 \times 10^{13}$

1205098_1102806_ans_148364609d754e27a91f81b809824923.JPG

An analyser is inclined to a polarizer at an angle of

$30^o$ . the intensity of light emerging from the analyser is

$\dfrac{1}{n}^{th}$ of that is incident on the polarizer. Then n is equal to

0%
$4$
0%
$\dfrac{4}{3}$
0%
$\dfrac{8}{3}$
0%
$\dfrac{1}{4}$

Explanation

Given,

$\theta=30^0$

Lets consider

$I_0=$ Intensity of incident light

Intensity of light passing though the analyser is given by

$I_a=\dfrac{I_0}{n}$ . . . . . .(1)

By malus law,

$I'=Icos^2 \theta$

$\dfrac{I_0}{n}=\dfrac{I_0}{2}cos^2 30^0$

$\dfrac{1}{n}=\dfrac{1}{2}\times (\dfrac{\sqrt{3}}{2})^2=\dfrac{3}{8}$

$n=\dfrac{8}{3}$

The correct option is C.

1467938_1096034_ans_d2894b4de05d4e138903347ffb86b0c0.png

Ratio of intensity of two waves is

$25 : 1$ . If interference occurs, then ratio of maximum and minimum intensity should be:-

0%
$25 : 1$
0%
$5 : 1$
0%
$9 : 4$
0%
$4 : 9$

The wavefront of a lightbeam is given by the equation

$x+2y+3z=c$ ,(where c is arbitary constant) the angle made by the direction of light with the y-axis is:

0%
${ cos }^{ -1 }\dfrac { 1 }{ \sqrt { 14 } }$
0%
${ cos }^{ -1 }\dfrac { 2 }{ \sqrt { 14 } }$
0%
${ sin }^{ -1 }\dfrac { 1 }{ \sqrt { 14 } }$
0%
${ sin }^{ -1 }\dfrac { 2 }{ \sqrt { 14 } }$

Explanation

Given,

Direction of light,

$\vec n=\hat i+2\hat j+3\hat k$

The angle made by the direction of light with y-axis is

$cos\beta=\hat n.\hat j$

$cos\beta=\dfrac{\vec n}{|\vec n|}.\hat j$

$cos\beta=\dfrac{(\hat i+2\hat j+3\hat k).(0\hat i+\hat j+0\hat k)}{\sqrt{(1)^2+(2)^2+(3)^2}}=\dfrac{2}{\sqrt{14}}$

$\beta=cos^{-1}\dfrac{2}{\sqrt{14}}$

The correct option is B.

How does the fringe width in a double slit interference pattern change, when the distance between the slits is increased ?

0%
Slit width decreases
0%
Slit width increases
0%
no effect
0%
Slits disappears

Explanation

We know that fringe width is equal to wavelength multiplied by distance from screen to slit divided by separation of two slit

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

Here,

Fring width is inversely proportional to slit separation therefore, When fring width will increase than slit separation will decease.

Correct option is A.

In a double slit experiment

$D=1m,d=0 .2cm$ and

$\lambda={ 6000 }\mathring { A }$ . The distance of the point from the central maximum where intensity is

$75\%$ of the at the centre will be:

0%
$0.01\ mm$
0%
$0.03\ mm$
0%
$0.05\ mm$
0%
$0.1\ mm$

Explanation

Given,

Distance between slit and screen is

$D=1m$

Slit width is

$d=0.2cm=0.2\times { 10 }^{ -2 }m$

Wavelength of light is

$\lambda =6000{ A }^{ 0 }=6000\times { 10 }^{ -7 }m$

$\frac { I }{ { I }_{ 0 } } =\frac { 75 }{ 100 } =\frac { 3 }{ 4 }$

Intensity of light in the fringe pattern is

$I={ I }_{ 0 }{ cos }^{ 2 }\left( \frac { \theta }{ 2 } \right)$

By equating

$\frac { 3 }{ 4 } { I }_{ 0 }={ I }_{ 0 }{ cos }^{ 2 }\left( \frac { \theta }{ 2 } \right) \\ \frac { \theta }{ 2 } =\frac { \pi }{ 12 } \\ \theta =\frac { \pi }{ 6 }$

Let

$\triangle \theta$ is path difference

$\frac { \pi }{ 6 } =\frac { \pi yd }{ \lambda D } \\ y=\frac { \lambda D }{ 6d }$

Substitute all value in above equation

$y=\frac { 6000\times { 10 }^{ -7 }\times 1 }{ 6\times 0.2\times { 10 }^{ -2 } } \\ \quad =0.05mm$

correct option is C.

Which of the following is incorrect?

0%
A thin convex lens of focal length ${f}_{1}$ is placed in contact with a thin concave lens of focal length ${f}_{2}$ . The combination will act as convex lens if ${f}_{1}<{f}_{2}$
0%
Light on reflection at water-glass boundary will undergo a phase change of $\pi$
0%
Spherical aberration is minimized by achromatic lens
0%
If the image of distant object is formed in front of the retina then defect of vision may be myopia

The two coherent sources of equal intensity produce maximum intensity of

$100$ units at a point. If the intensity of one of the sources is reduced by

$50\%$ by reducing its width then the intensity of light at the same point will be

0%
90
0%
81
0%
67
0%
72.85

Explanation

Hint: The resultant intensity of sources can be written as,

$I_{ R }={ I }_{ 1 }+{ I }_{ 2 }+2\sqrt { { I }_{ 1 }{ I }_{ 2 } }$

Step 1: Given data

Maximum intensity

${ I }_{ 0 }=100units$

Intensity of one source is reduced by

$50 percent$

We know that when two source having intensity of

$I_{ 1 }\quad and\quad { I }_{ 2 }$ Then resultant intensity is

$I_{ R }={ I }_{ 1 }+{ I }_{ 2 }+2\sqrt { { I }_{ 1 }{ I }_{ 2 } }$

Here,

${ I }_{ 1 }={ I }_{ 0 }\\ { I }_{ 2 }={ I }_{ 0 }$

Putting all value in above equation

$I_{ R }={ I }_{ 0 }+ { I }_{ 0 } +2\sqrt { { I }_{ 0 }\times { I }_{ 0 } } \\ \quad ={ I }_{ 0 }(1+1 +2 )\\$

${ I }_{ 0 }=\frac { { I }_{ R } }{ 4 } =\frac { 100 }{ 4 } =25$

Step 2: Find the resultant intensity.

When one source is reduced

${ I }_{ 1 }={ I }_{ 0 }\\ { I }_{ 2 }={ I }_{ 0 }-{ I }_{ 0 }\times \frac { 50 }{ 100 } =\frac { { I }_{ 0 } }{ 2 }$

Putting all value in above equation

$I_{ R }={ I }_{ 0 }+\frac { { I }_{ 0 } }{ 2 } +2\sqrt { { I }_{ 0 }\times \frac { { I }_{ 0 } }{ 2 } } \\ \quad ={ I }_{ 0 }(1+\frac { 1 }{ 2 } +\sqrt { 2 } )\\ \quad =25\times 2.91\\ \quad =72.85$

Correct option is D.

The speed of the light in the medium is :-

0%
maximum on the axis of the beam
0%
minimum on the axis of the beam
0%
the same everywhere in the beam
0%
directly proportional to the intensity I

Monochromatic green light of wavelength

$5\times10^{-7}m$ illuminates a pair of slits 1 mm apart. The separation of bright lines on the interference pattern formed on a screen 2 m away is

0%
1.0 mm
0%
1.5 mm
0%
2.0 mm
0%
1.9 mm

Explanation

We know that for Young's Double Slit Experiment

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

Here,

$\lambda=5\times10^{-7}m^{-1}\\D=2m\\d=10^{-3}m$

Now,

$\beta=\cfrac{5\times10^{-7}\times2}{10^{-3}}m=10^{-3}=1.0mm$

Thus

$\beta=1.0mm$

Two superimposing waves are represented by equation

${y}_{1}=2\sin { 2\pi } \left( 10t-0.4x \right)$ and

${y}_{2}=4\sin { 2\pi } \left( 20t-0.8x \right)$ . The ratio of

${I}_{max}$ to

${I}_{min}$ is

0%
$36:4$
0%
$25:9$
0%
$1:4$
0%
$4:1$

In a Young's double slit experiment, constructive interference is produced at a certain point

$P$ . The intensities of light at

$P$ due to the individual sources are

$4$ and

$9$ units. The resultant intensity at point

$P$ will be-

0%
$13$ units
0%
$25$ units
0%
$\sqrt{97}$ units
0%
$5$ units

Explanation

$I_t = \left(\sqrt{I}_1 + \sqrt{I}_2\right)^2$

$= (2 + 3)^2$

$= 25$

After reflection from a concave mirror, a plane wave front becomes

0%
Cylindrical
0%
Spherical
0%
Remains planar
0%
None of the above

When waves of same intensity from two coherent sources reach a point with zero path different the resulting intensity is K . When the above path difference is

$\lambda /4$ the intensity becomes

0%
K
0%
K/2
0%
K/4
0%
K/8

Explanation

The maximum intensity is formed at

$\Delta x=0$ which is

$k$

Intensity at any other point having path difference

$\Delta x$ is given by

$k'=k\cos ^{ 2 }{ \left( \cfrac { \pi }{ \lambda } \Delta x \right) }$

For

$\Delta x=\cfrac{\lambda}{4}$ ,

$\cfrac{\pi}{\lambda}.\Delta x=\frac{\pi}{4}$ so

$k'=k\cos ^{ 2 }{ \left( \cfrac { \pi }{ 4 } \right) } =\cfrac{k}{4}$

$s_1$ and

$s_1$ are two sources of sound emitting sine waves. The two sporces are in phase. The source emittied by the two sources interfere at point F. The waves of wavelength:

0%
1 m will result in constructive interference
0%
$\frac { 2 }{ 3 } m$ will result in constructive interference
0%
2 m will result in destructive interference
0%
4 m will result in destructive interference

Explanation

We know that,

For constructive

Path difference = $n λ$

${{S}_{1}}F-{{S}_{2}}F=n\lambda$

$6m-4m=n\lambda$

$2\,m=n\lambda$

$\lambda =\dfrac{n}{2}\,m$

Where, $n= 0, 1, 2….$

So, $\lambda =0,0.5\,m,1\,m,\dfrac{3}{2}\,m,2\,m....$

Hence, the wave length is $1\ m$ in constructive interference

Direction of the first secondary maximum in the fraunhoffer diffraction pattern at a single slit is given by:

0%
a sin $\theta =\frac{\lambda }{2}$
0%
a sin $\theta =\frac{3\lambda }{2}$
0%
acos $\theta =\frac{3\lambda }{2}$
0%
a sin $\theta =\lambda$

In the interference, two light waves are in the phase difference of at a point. If the yellow light is used, then the color of fringes at that point will be:-

0%
Yellow
0%
whitw
0%
red
0%
Black

The distances of interference point on a screen from two slits are

$18 \, \mu \, m$ and 12.3

$\mu$ m. If the wavelength of light used is

$6 \times 10^{-7}m$ then the number of dark or bright fringe formed there will be -

0%
$8^{th}$ dark
0%
$9^{th}$ bright
0%
$10^{th}$ dark
0%
$11^{th}$ dark

Explanation

$\begin{array}{l} \Delta =d\sin \theta \\ 5.7\mu m=d\theta \\ \theta =\dfrac { { 5.7\mu n } }{ d } \\ x=\dfrac { { \Delta D } }{ d } \\ \dfrac { \Delta }{ x } =\dfrac { { 5.7 } }{ { 6\times { { 10 }^{ -7 } } } } \\ \dfrac { \Delta }{ x } =\dfrac { { 56 } }{ 6 } =\dfrac { { 19 } }{ 2 } \\ Hence,\, we\, have \\ \dfrac { \Delta }{ x } =9\dfrac { 1 }{ 2 } \end{array}$

In an experiment the two slits are

$0.5$ mm apart and the fringes are observed to

$100$ cm from the plane of the slits. The distance of the

$11th$ bright fringe from the 1st bright fringe is

$9.72mm$ . Calculate the wavelength-

0%
$4.86 \times {10^{ - 5}}cm$
0%
$4.86 \times {10^{ - 8}}cm$
0%
$4.86 \times {10^{ - 6}}cm$
0%
$4.86 \times {10^{ - 7}}cm$

Explanation

We have,

$d = 0.5$ mm

$D = 100 \ cm = 1000 \ mm$

we know that,

path difference

$(\triangle x)$

$\triangle x= \dfrac{xd}{D}$

for maxima (bright triangles)

$\triangle x = n\lambda$

where

$\lambda \rightarrow$ wavelength

$\Rightarrow \dfrac{xd}{D} = n\lambda$

$\Rightarrow x = \dfrac{n\lambda D}{d}$

now,

distance

$(x_{1})$ & I bright fringe

putting

$n = 1$

$x_{1} = \dfrac{\lambda D}{d}$

distance

$(x_{11})$ of

$11^{th}$ bright fringe (putting

$n = 11$ )

$x_{11} = \dfrac{11\lambda D}{d}$

according to question

$x_{11}-x_{1} = 9.72 mm$

$\Rightarrow \dfrac{11\lambda D}{d}-\dfrac{\lambda D}{d} = 9.72$

$\Rightarrow \dfrac{10\lambda 0}{d} = 9.72$

$\Rightarrow \lambda = \dfrac{9.72\times d}{10D}$

$=\dfrac{9.72\times 0.5}{10\times 1000}$

$\Rightarrow \lambda = 4.86\times 10^{-4}mm$

$\lambda = 4.86\times 10^{-5}cm$

1382152_1212630_ans_53194c334b9540f1a484382080be7abe.JPG

CBSE Questions for Class 12 Medical Physics Wave Optics Quiz 9 - 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 9 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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