produced by a source at infinity

produced by a source at finite distance

MCQ Questions for Class 12 Medical Physics Wave Optics Quiz 10

Angular width of central maximum of a diffraction pattern on a single does not depend upon :

0%
Distance between slit and source
0%
Wavelength of the light used
0%
width of the slit
0%
None of the above

In Young's double slit experiment

$10th$ order maxima is obtained at the point of observation in the interference pattern for

$\lambda = 7000 A^o$ . If the source is replaced by another one of wavelength

$5000\ A^o$ then the order of maximum at the same point will be

0%
$12\ th$
0%
$14\ th$
0%
$16\ th$
0%
$18\ th$

A parallel beam of monochromatic light of wavelength 5000 A is incident on a single narrow slit of width 0.001 mm. The light is focussed by a convex lens on a screen placed on focal plane. The first minimum will be formed formed for the angle of diffraction equal to

0%
$20^{\circ}$
0%
$15^{\circ}$
0%
$30^{\circ}$
0%
$50^{\circ}$

Explanation

$d\sin\theta =m\lambda$

$\theta =\sin^{-1}\left(\dfrac{m\lambda}{d}\right)$

$=\sin^{-1}\left(\dfrac{(1)(5000\times 10^{-10})}{0.001\times 10^{-3}}\right)$

$\theta =30^o$ .

Two coherent source must have the same :

0%
amplitude
0%
phase difference
0%
frequency
0%
both B and C

A monochromatic light is incident on a single slit of width

$24\times 10^{-5}\ cm$ . Calculate wavelength of light if angular position of

$1^{st}$ secondary maxima is

$30^{o}$ .

0%
$8000\ A^{o}$
0%
$12000\ A^{o}$
0%
$6000\ A^{o}$
0%
$16000\ A^{o}$

A slit of size

$0.15cm$ is placed at

$2.1m$ from a screen. On illuminated it by a light of wavelength

$5 \times {10^{ - 5}}cm$ . The width of diffraction pattern will be:-

0%
$70nm$
0%
$0.14nm$
0%
$1.4nm$
0%
$.014nm$

A pale wavefront of wavelength

$\lambda$ is incident in a single slit of width a

0%
$\dfrac {\lambda} {a}$
0%
$\dfrac {2 \lambda} {a}$
0%
$\dfrac {a} {\lambda}$
0%
$\dfrac {a} {2 \lambda}$

The first diffraction minimum due to single slit diffraction is

$\theta ,$ for a light of wavelength

$5000 A.$ If the width of the sit is

$1 \times {10^{ - 4}},$ then the value of

$\theta$ is

0%
${30^0}$
0%
${45^0}$
0%
${60^0}$
0%
${15^0}$

Explanation

$\begin{array}{l} d\sin \theta =n\lambda ,n=1 \\ 1\times { 10^{ -6 } }\times \sin \theta =5000\times { 10^{ -10 } } \\ \sin \theta =\frac { 1 }{ 2 } \\ \theta =30 \end{array}$

Two identical coherent sources placed on a diameter of a circle of radius R at separation

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

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

$\left( {x = 5\lambda } \right)$

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

In an interference pattern of two waves fringe width is

$\beta$ . If the frequency of source is doubled then fringe width will become:-

0%
$\dfrac{1}{2} \beta$
0%
$\beta$
0%
$2\beta$
0%
$\dfrac{3}{2} \beta$

Explanation

Fringe width is given by,

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

Where, $\lambda$ is the wavelength of light, $D$ is the distance from slit to screen and $d$ is the slit width.

We know,

$c=\lambda v$

$c$ is speed of light and $ν$ is frequency of light.

If $ν$ is doubled, then,

$c=\lambda' 2v=\lambda v$

$\implies \lambda'=\dfrac{\lambda}{2}$

So, the fringe width becomes,

$\beta'=\dfrac{\lambda' D}{d}=\dfrac 12 \dfrac{\lambda D}{d}=\dfrac{\beta}{2}$

Thus, the fringe width becomes half of the original value when the frequency of light is doubled.

$Y.D.S.E$ . If the width of the slits are gradually decreased, then

0%
Bright fringe will become brighter and dark fringe become darker
0%
Bright fringe become less bright and dark fringe becomes less dark
0%
Bright fringe become brighter and dark fringe become lighter
0%
Bright fringe become less bright and dark fringe becomes less darker

A light source of diameter

$2 cm$ is placed

$20 cm$ behind a circular opaque disc to diameter

$4 cm$ . Shadow is formed on a screen at a distance of

$80 cm$ , the ratio of the area of umbra and penumbra shadow region is equal to

0%
$0.58$
0%
$0.22$
0%
$0.18$
0%
$0.11$

Angle of incidence is equal to the angle of reflection.

0%
Always
0%
Sometimes
0%
Under special condition
0%
Never

Explanation

As per the low of reflection the angle of incident is always equal to angle

of reflection.

Hence option

$A$ is correct answer.

The wave theory of light was given by

0%
HUygen
0%
Young
0%
Maxwell
0%
Olank

Explanation

$\textbf{Wave theory of light given by Huygen.}$

In YDSE, d = 2 mm, D = 2 m and

$\lambda$ = 500 nm. If intensity of two slits are

$l _ { 0 }$ and

$9 l _ { 0 }$ then find intensity at

$y = \frac { 1 } { 6 }$

0%
$7 l _ { 0 }$
0%
$10 l _ { 0 }$
0%
$16 l _ { 0 }$
0%
$4 l _ { 0 }$

Explanation

Given

$:$

$d=2mm$

$D=2m$

lambda

$=500nm$

Intensities

$:$

${l_0}$ and

$9{l_0}$

path differences

$=xd/D$

$\frac{1}{6} \times {10^{ - 3}} \times 2 \times {10^{ - 3}}/2$

$= 1.66 \times {10^{ - 7}}m$

Phase difference

$= \frac{{2\pi \Delta x}}{\lambda }$

$= 2.08 = 0.664\pi = {119.52^0}$

therefore

$,$

${I_{net}} = {l_0} + 9{l_0} + 2{\left( {9{l_0}} \right)^{1/2}}\cos \left( {119.52} \right)$

$= 10{l_0} - 2.9{l_0}$

$= 7.04{l_0} \approx 7{l_0}$

Hence,

option

$(A)$ is correct answer.

The displacement of the interfering sound waves are

${ y }_{ 1 }=4sincot$ and

${ y }_{ 2 }=3sin\left( cot+\frac { \pi }{ 2 } \right)$ . What is the amplitude of the resultant wave

0%
5
0%
7
0%
1
0%
0

Light of wavelength

$\lambda$ is incident on a slit of width d. The resulting Diffraction pattern is observed on screen at a distance D. the linear width of the principle maximum is then equal to width of the slit if D equals :

0%
$\dfrac {d} {\lambda}$
0%
$\dfrac {2\lambda} {d}$
0%
$\dfrac{{ d }^{ 2 }} {2\lambda}$
0%
$\dfrac {2\lambda ^{ 2 }} {d}$

Statement-1 : In standard YDSE set up with visible light, the position on screen where phase difference is zero appears bright.
and
Statement-2 : In YDSE set up magnitude of electromagnetic field at central bright fringe is not varying with time.

0%
Statement-1 is true statement-2 is true and statement-2 is correct explanation for statement-1.
0%
Statement-1 is true statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
0%
Statement-1 is true statement-2 is false.
0%
Statement-1 is false statement-2 is true.

In a diffraction pattern due to a single slit of width

$a$ , the first minimum is observed at an angle

$30^{\circ}$ when light of wavelength

$5000\overset {\circ}{A}$ is incident on the slit. The first secondary maximum is observed at an angle of

0%
$\sin^{-1} \left (\dfrac {3}{4}\right )$
0%
$\sin^{-1} \left (\dfrac {1}{4}\right )$
0%
$\sin^{-1} \left (\dfrac {2}{3}\right )$
0%
$\sin^{-1} \left (\dfrac {1}{2}\right )$

If light waves emitted by an ordinary source, for what period of time, the phase remains constant :-

0%
$10 sec$
0%
$1 sec$
0%
$10^{-3}$ sec
0%
$10^{-8}$ sec

$I_0$ 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%
$I_0$
0%
$\dfrac{I_0}{2}$
0%
$2I_0$
0%
$4I_0$

Explanation

$(I_0)$ Maximum independent of slit width. Thus intensity will remains unchanged.

The main difference in the phenomenon of interference and diffraction is that:-

0%
diffraction is due to interaction of light from the same wavefront whereas interference is the interaction of (light) waves from two isolated source.
0%
diffraction is due to interaction of light from same wavefront whereas interference is the interaction of two waves derived from the same source.
0%
diffraction is due to interaction of waves derived from the same source, whereas the interference is the bending of light from the same waveform
0%
diffraction is caused by the reflected waves from a source whereas interference is caused due to waves from a source

Two point source separated by

$d=5\mu m$ emit light of wavelength

$\lambda=2\mu m$ in phase. A circular wire of radius

$20\ mu m$ is placed around the source as shown in figure.

0%
Points $A$ and $B$ are dark and points $C$ and $D$ are bright.
0%
Points $A$ and $B$ are bright and points $C$ and $D$ are dark.
0%
Points $A$ and $C$ are dark and points $B$ and $D$ are bright.
0%
Points $A$ and $C$ are bright and points $B$ and $D$ are dark.

A plane wave front of wave length

$6000A$ is incident upon a slit of

$0.2mm$ width, which enables fraunhofer's diffraction pattern to be obtained on a screen

$2m$ away. Width of the central maxima in mm will be

0%
$10$
0%
$12$
0%
$8$
0%
$2$

Explanation

Given,

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

$d=0.2mm$

$D=2m$

Fringe width,

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

$\beta=\dfrac{6000\times 10^{-10}\times 2}{0.2\times 10^{-3}}=6mm$

Width of the central maxima,

$W=2\beta=2\times 6=12mm$

The correct option is B.

Parallel beam is

0%
produced by a source at infinity
0%
produced by a source at finite distance
0%
produced by a virtual source
0%
None of these

What is the path difference of destructive interference :

0%
$n\lambda$
0%
$n(\lambda +1)$
0%
$\cfrac{(n+1)\lambda}{2}$
0%
$\cfrac{(2n+1)\lambda}{2}$

Explanation

For destructive interference

Intensity should be minimum

$I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \phi$

$\Rightarrow \cos \phi = -1 = \left(\dfrac{2n + 1}{2}\right) \pi = \phi$

$\Rightarrow \Delta x = \left(\dfrac{2n + 1}{2} \right) \lambda$

In diffraction using single slit, a slit of width a is illuminated by white light. For red light

$\left( \lambda =6500\quad \overset { \circ }{ A } \right)$ , the first minima is obtained for

$\theta ={ 30 }^{ \circ }$ . Then the value of a will be:

0%
3250 $\overset { \circ }{ A }$
0%
$6.5\times { 10 }^{ -4 }$
0%
1.24 $\mu m$
0%
$2.6\times { 10 }^{ -4 }$

Explanation

For first minima

$\theta=\dfrac{\lambda}{a}$ or

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

$\therefore a=\dfrac{6500 \times 10^{-8} \times 6}{\pi}\ (As\ 30^o=\dfrac{\pi}{6}\ radiant)$

$=1.24 \times 10^{-4}\ cm=1.24\mu m$

Newton's corpuscular model of light.

0%
successfully explain the rectilinear propagation of light, reflection of light and refraction of light
0%
does not explain the phenomena of interference of light, diffraction of light and polarisation of light
0%
is based upon particle theory of light
0%
None of the above

In the case of parallel beam,

0%
intensity gradually increases
0%
intensity gradually decreases
0%
intensity remains constant
0%
None of these

According to Huygen's theory of light

0%
light is stream of photon
0%
light is wave
0%
light behaves as particle
0%
None of the above

What will be the angular width of central maxima in Fraunhoffer diffraction when light of wavelength

$6000\ \mathring {A}$ is used and slit width is

$12\ \times 10^{-5}\ cm$ .

0%
$2\ rad$
0%
$3\ rad$
0%
$1\ rad$
0%
$8\ rad$

Explanation

Given,

$\lambda=6000\mathring{A}$

$d=12\times 10^{-5}cm$

The angular width of central maxima in Fraunhoffer diffraction is given by,

full angular width

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

$=\dfrac{2 \times 6000 \times 10^{-10}}{12 \times 10^{-5}}$

$=1rad$

The bending of beam of light around corners of obstacles is called

0%
Reflecton
0%
Diffraction
0%
Refraction
0%
Interference

In a biprism experiment, the distance of 20 th bright bandfrom the center of the interference pattern is 8

$\mathrm { mm }$ . The distance of 30th bright band from the center is

0%
$11.8\mathrm { mm }$
0%
12 $\mathrm { mm }$
0%
14 $\mathrm { mm }$
0%
16 $\mathrm { mm }$

Explanation

Given

$\begin{array}{l} 20\beta =8mm \\ \beta =\dfrac { 8 }{ { 20 } } \\ Now, \\ 30th\, \, \max ima=30\beta =\dfrac { { 30\times \beta } }{ { 20 } } =12mm \\ 30th\, \min ima=\dfrac { { \left( { 2\left( { 30 } \right) -1 } \right) } }{ 2 } \beta \\ =\dfrac { { 59 } }{ 2 } \beta \\ =\dfrac { { 59 } }{ 2 } \times \dfrac { 8 }{ { 20 } } \\ =11.8mm \\ Hence,\, option\, A\, is\, the\, correct\, answer. \end{array}$

If the ratio of maximum and maximum Intensities tn an interference pattern is 36: 1 then the ratio of amplitudes of two Interfering waves Will be

0%
5 :7
0%
7 :4
0%
4 :7
0%
7 :5

Explanation

$\frac{{{I_{\max }}}}{{{I_{\min }}}} = \left( {\frac{{\frac{{{a_1}}}{{{a_2}}} + 1}}{{\frac{{{a_1}}}{{{a_2}}} - 1}}} \right)$

$\Rightarrow \frac{{{a_1} + {a_2}}}{{{a_1} - {a_2}}} = 6\frac{{{a_2}}}{{{a_1}}} = 7:5$

Hence,

option

$(D)$ is correct answer.

In Y.D.S.E. how many maxima can be obtained on the screen if wave length of light used is

$200 mm$ &

$d = 700 mm$ :-

0%
$12$
0%
$7$
0%
$18$
0%
$14$

What is the effect on the interference fringes in Young's double slit experiment if the source slit is moved closer to the double slit plane?

0%
The fringe width increases
0%
The fringe width decreases
0%
The fringes become more distinct
0%
The fringes become less distinct

If the frequency of the source is doubled in Young's double slit experiment, then fringe width

$( \beta )$ will be-

0%
unchanged
0%
$\beta / 2$
0%
$2 \beta$
0%
$3 \beta$

Explanation

$\begin{array}{l} fringe\, \, width\, \, \beta =\frac { { \lambda D } }{ \alpha } \\ And\, \, \, \lambda =\frac { c }{ f } \\ so,\, \, \, \, \, \beta =\frac { { CD } }{ { fd } } \\ Hence,\, \, \beta \propto \frac { 1 }{ f } \end{array}$

If frequency double, fringe width

$(\beta)$ is halved

Option

$B$ is correct.

In YDSE if white is used then.

0%
except center, there will be spectrum
0%
except center no spectrum any where
0%
spectrum every where
0%
spectrum at center only

In a double-slit experiment, green light

$(5303\overset{o}{A})$ falls on a double slit having a separation of

$19.44\ \mu m$ and a width of

$4.05\ \mu m$ . The number of bright fringes between the first and the second diffraction minima is?

0%
$09$
0%
$10$
0%
$04$
0%
$05$

Explanation

For diffraction
location of

$1^{st}$ minima

$y_1=\dfrac{D\lambda}{a}=0.2469D\lambda$
Location of

$2^{nd}$ minima

$y_2=\dfrac{2D\lambda}{a}=0.4938D\lambda$
Now for interference
Path difference at P.

$\dfrac{dy}{D}=4.8\lambda$
path difference at Q

$\dfrac{dy}{D}=9.6\lambda$
So orders of maxima in between P & Q is

$5, 6, 7, 8, 9$
So

$5$ bright fringes all present between P & Q.
1142418_1331389_ans_46ce40a1b37f49a69322f730edda95e6.PNG

When an unpolarized light of intensity

$I_0$ is incident on a polarizing sheet the intensity of the light which does not get transmitted is?

0%
Zero
0%
${I}_{0}$
0%
$\cfrac{1}{2}{I}_{0}$
0%
None

Huygen's theory of secondary waves can be used of find-

0%
Velocity of light
0%
The wavelength of light
0%
Wave front geometrically
0%
Magnifying power of microscope

The wavefront of a light beam is given by the equation

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

0%
$\cos ^ { - 2 } \frac { 2 } { \sqrt { 14 } }$
0%
$\cos ^ { - 1 } \frac { 2 } { \sqrt { 14 } }$
0%
$\cos ^ { - 3 } \frac { 2 } { \sqrt { 14 } }$
0%
$\cos ^ { - 4 } \frac { 2 } { \sqrt { 14 } }$

Explanation

The wave front equations must be of form

$ax+by+cz=1$ where

${a^2} + {b^2} + {c^2} = 1$

$($

$a,b,c$ are cosines of angles made with

$x,yx,z$ axes respectively

$)$

applying this to the given equation

$:$

${1^2} + {2^2} + {3^2} = {c^2}$

${c^2} = 14$

$c = \sqrt {14}$

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

angle made with

$y-$ axis is

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

Hence,

option

$(B)$ is correct answer.

Consider Fraunhoffer diffraction pattern obtained with a single slit illuminate incidance. At the angular position of the first diffraction minimum in the phase( radian) between the wavelets from the opposite edges of the slit is

0%
$\dfrac {\pi} 4$
0%
$\dfrac {\pi} 2$
0%
${\pi}$
0%
$2\pi$

What wavelength of light would you be able to resolve at the fastest distance based on diffraction?

0%
Red $(\lambda=650nm)$
0%
Green $(\lambda=550nm)$
0%
Blue $(\lambda=450nm)$
0%
All the wavelengths will be resolved equally

Fraunhoffer lines are obtained in

0%
Solar spectrum
0%
The spectrum obtained from neon lamp
0%
Spectrum from a discharge tube
0%
None of the above

The figure shows a double slit experiment where

$P$ and

$Q$ are the slits. The path lengths

$P X$ and

$Q X$ are

$n \lambda$ and

$(n+2) \lambda$ respectively, where

$n$ is a whole number and

$\lambda$ is the wavelength. Taking the central fringe as zero, what is formed at

$X$

0%
First bright
0%
First dark
0%
Second bright
0%
Second dark

Explanation

For brightness path difference

$=n \lambda = 2\lambda$
So second is bright.

Which of following nature of light waves is supported by the phenomenon of interference:

0%
longitudinal
0%
transverse
0%
both transverse and longitudinal
0%
none of the above

Light of wavelength 6000 is incident bria single slit. The first minimum of the diffraction pattern is obtained at 4 mm from the centre. The screen is at a distance of 2 m from the slit.

0%
0.15 mm
0%
0.1 mm
0%
0.3 mm
0%
0.2 mm

We could observe path of light due to ........... of light from tiny particle of solution of transparent medium.

0%
Scatterign
0%
Dispersion
0%
Refraction
0%
Reflaction

Electrons accelerated by potential

$V$ are diffracted from a crystal. If

$d = 1\overset {\circ}{A}$ and

$i = 30^{\circ}, V$ should be about

$(h = 6.6\times 10^{-34} J-s, m_{e} = 9.1\times 10^{-31} kg, e = 1.6\times 10^{-19}C)$ .

0%
$2000\ V$
0%
$50\ V$
0%
$500\ V$
0%
$1000\ V$

CBSE Questions for Class 12 Medical Physics Wave Optics Quiz 10 - 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 10 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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