Match the following : PART-A PART-B a) Polarisation e) All types of waves b) interference f) longitudinal waves c) diffraction g) transverse waves d) reflection h) only with transverse waves i) stationary waves produced in stretched strings

a$$\rightarrow$$ g ; b$$\rightarrow$$ e, f , g,i ; c$$\rightarrow$$ e, f , g ; d $$\rightarrow$$ e, f , g

a$$\rightarrow$$h, g; b$$\rightarrow$$ f , g ; c$$\rightarrow$$ g ; d $$\rightarrow$$ h

a$$\rightarrow$$ e, f , g; b$$\rightarrow$$ g ; c$$\rightarrow$$ e, f , g ; d $$\rightarrow$$ g

a$$\rightarrow$$ e; b$$\rightarrow$$ h,i ; c$$\rightarrow$$ g, h ; d $$\rightarrow$$ e

MCQ Questions for Class 12 Medical Physics Wave Optics Quiz 4

Ramesh was playing in the garden on a sunny day. He observed almost elliptical splashes of sunlight on the ground as he walked under a tree. What did he observe ?

0%
Shadow of the leaves
0%
Shadow of the fruits on the tree
0%
Pinhole-camera images of sun formed by spaces between leaves in canopy.
0%
None of these

Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil of diameter 3 mm. Approximately what is the maximum distance up to which these dots can be resolved by the eye.

0%
5 m
0%
6 m
0%
1 m
0%
4 m

The first diffraction minimum due to a single slit Fraunhoffer diffraction is at the angle of diffraction

$30 ^ { \circ }$ for a light of wave length 5000

$A ^ { 0 }$ .The width of the slit is

0%
$1.0 \times 10 ^ { - 4 } \mathrm { cm }$
0%
$2.164 \times 10 ^ { - 4 } \mathrm { cm }$
0%
$1.082 \times 10 ^ { - 3 } \mathrm { cm }$
0%
$0.546 \mathrm { cm }$

Explanation

$\begin{array}{l} According\, to\, question............... \\ For\, \, Diffraction\, \, minima= \\ d\sin \theta =n\lambda \\ 1st\, order\, n=1 \\ \Rightarrow d\, sin30=1\times 5000\times { 10^{ -10 } } \\ \Rightarrow d\times \frac { 1 }{ 2 } =5000\times { 10^{ -10 } } \\ \Rightarrow d=10000\times { 10^{ -10 } }m \\ \, \, \, \, \, \, \therefore \, \, \, \, d=1\, \times { 10^{ -4 } }m \\ so\, that\, the\, correct\, option\, is\, A. \end{array}$

Two light sources are coherent when:

0%
Only their frequency are equal
0%
their wavelength are equal
0%
Their amplitudes are equal
0%
Time frequency are equal and their phase

Explanation

$Given:$ Two light sources are coherent when

$Solution:$ If two independent sources emitting light of the same wavelength are said to be coherent.

$So,the$

$correct$

$option:B$

The diffraction effect can be observed in

0%
only sound waves
0%
only light waves
0%
only ultrasonic waves
0%
sound as well as light waves

A double slit is illuminated by light of wave length 6000

$A ^ { 0 }$ .The slits are 0.1 cm apart and the screen is placed 1 m away. Then the angular position of 10th maxima is

0%
$6 \times 10 ^ { - 3 } r a d$
0%
6 rad
0%
$0.006 ^ { \circ }$
0%
$6 ^ { 0 }$

Explanation

$\begin{array}{l} \lambda =6000{ { A }^{ o } } \\ d=0.1cm \\ D=1 \\ angular=\frac { { 10\beta } }{ D } \\ =\frac { { 10\times \lambda D } }{ { dD } } =\frac { { 10\lambda } }{ d } =\frac { { 10\times 6000\times { { 10 }^{ -10 } }\times 100 } }{ { o.1 } } \\ =6\times { 10^{ -3 } } \\ =6rad \\ Hence, \\ option\, \, B\, \, is\, correct\, \, answer. \end{array}$

The angular spread of central maximum, in the diffraction pattern, does not depend on ______

0%
the distance between the slit and sources
0%
width of slit
0%
wavelength of light
0%
frequency of light

Explanation

Angular spread of central maxima

$=(\theta)=\cfrac{\lambda}{b}$

Where

$\lambda=$ wavelength of light

$b=$ width of slit

$\Rightarrow$ Clearly, it does depends on the distance between slit and sources.

Which of the following is correct for light diverging from a point source ?

0%
The intensity decreases in proportion for the distance squared.
0%
The wavefront is parabolic.
0%
The intensity at the wavelength does not depend on the distance.
0%
None of these.

Explanation

A point source produces a spherical wavefront that moves in all the directions from the point.

Whereas the intensity at the wavelength is dependent on the distance and it follows the inverse square law. So, the intensity decreases with an increase in the distance from the source.

Thus, option

$(A)$ is correct.

To observe diffraction, the size of the obstacle

0%
Should be $\lambda /2$ where 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.

Which one is not produced by sound waves in air?

0%
Polarisation
0%
Diffraction
0%
Refraction
0%
Reflection

Coherent sources are characterized by the same:

0%
phase and phase velocity
0%
wavelength, amplitude and phase velocity
0%
wavelength, amplitude and frequency
0%
wavelength and phase

The wavelength of light visible to eye is of the order of

0%
$10^{-2}\ m$
0%
$10^{-10}\ m$
0%
$1\ m$
0%
$6\times 10^{-7}\ m$

Explanation

The range for visible light is between

$4 \times 10^{-7} m$ to

$7\times 10^{-7} m$ . The only wavelength that is present in this range given as an option is

$6\times 10^{-7}m$ .

Hence, the correct answer is option (D).

Which of the following phenomenon can explain quantum nature of light

0%
Photoelectric effect
0%
Interference
0%
Diffrection
0%
Polarisation

$L$ is the coherent length and

$c$ the velocity of light, the coherent time is

0%
$cL$
0%
$\dfrac{L}{c}$
0%
$\dfrac{c}{L}$
0%
$\dfrac{1}{Lc}$

Explanation

Coherent time

$=\dfrac{Coherence\ length}{Velocity\ of\ light}=\dfrac{L}{c}$

A star moves away from earth at speed

$0.8\ c$ while emitting light of frequency

$6\times 10^{14}Hz$ . What frequency will be observed on the earth (in units of

$10\ Hz$ )(

$c=$ speed of light)

0%
$0.24$
0%
$1.2$
0%
$30$
0%
$3.3$

Explanation

Observed frequency

$c'=c\left(1-\dfrac{v}{c}\right)$

$\Rightarrow v'=6\times 10^{14}\left(1-\dfrac{0.8\ c}{c}\right)$

$=1.2\times 10^{14}Hz$

The wave front due to a source situated at infinity is :

0%
spherical
0%
cylindrical
0%
planar
0%
none of these

In Young's double slit experiment, first slit has width four times the width of the second slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe system is:

0%
$2:1$
0%
$4:1$
0%
$9:1$
0%
$8:1$

Explanation

Let intensity from the smaller slit be

$I$ so, the intensity from the bigger slit will be

$4I$ since it is 4 times of smaller slit.
Now,

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

$=5I+2\sqrt{4I^{2}}$

$=5I+2(2I)$

$=9I$

$Imin=4I+I-2\sqrt{I(4I)}$

$=5I-2(2I)$

$=I.$
So,

$\dfrac{I_{max}}{I_{min}}=\dfrac{9I}{I}=\dfrac{9}{1}.$

In Young's double slit experiment, the two equally bright slits are coherent, but of phase difference

$\dfrac{\pi }{3}$ . If maximum intensity on the screen is

$I_{o}$ , the intensity at the point on the screen equidistant from the slits is:

0%
$I_{o}$
0%
$\dfrac{I_{o}}{2}$
0%
$\dfrac{I_{o}}{4}$
0%
$\dfrac{3I_{o}}{4}$

Explanation

Given

$:\phi =\pi /3$

$I=I_{max}cos^{2}(\dfrac{\phi }{2})$

$I=I_{0}cos^{2}(\dfrac{\pi /3}{2})$

$=I_{0}cos^{2}(\pi /6)$

$=I_{0}(\dfrac{\sqrt{3}}{2})^{2} (\because cos\pi /6=\dfrac{\sqrt{3}}{2})$

$=I_{0}\dfrac{3}{4}$

So, intensity on the screen equidistant from the slats is

$\dfrac{3}{4}I_{0}.$

Light of wavelength

$6000\ A$

$^{\circ}$ from a distant source falls on a slit

$0.5\ mm$ wide. The distance between two dark bands on each side of the central bright band of the diffraction pattern observed on a screen placed at a distance

$2\ m$ from the slit is:

0%
$1.2\ nm$
0%
$2.4\ nm$
0%
$3.6\ nm$
0%
$4.8\ mm$

Explanation

Distance between 2 dark bands on each side of the central bright band

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

$=\dfrac{2\times 6000\times 10^{-10}\times 2}{0.5\times 10^{-3}}$

$=4.8\times 10^{-3}m$

$=4.8\ mm.$

Anti-nodal curves represent the points joining:

0%
destructive interference
0%
constructive interference
0%
equal phase curves
0%
equal pressure curves
0%
zero pressure curves

Explanation

Conceptual ,
anti node represents joining of all points in

$constructive$

$interface$ as the

$amplitude$ is

$high$
option

$B$ is correct

In a Young's double slit experiment using red and blue lights of wavelengths

$600\ mm$ and

$480\ nm$ respectively, the value of

$n$ for which the

$n^{th}$ red fringe coincides with

$(n + 1)^{th}$ blue fringe is

0%
$5$
0%
$4$
0%
$3$
0%
$2$

If one of the two slits of a Young's double slit experiment is painted over so that it transmits half the light intensity of the other, then:

0%
the fringe system would disappear
0%
the bright fringes would be brighter & dark fringes would be darker
0%
the dark fringes would be brighter and bright fringes would be darker
0%
bright as well as dark fringes would be darker

Explanation

Intensity

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

$I_{2} =2 I_{1}$

Maximum intensity will be at

$\phi = 0$ and minimum at

$\phi = \pi/2$

$I_{max} = I_{1} + I_{2} + 2I_{1}I_{2} = I_{1} +2 I_{1} +2\sqrt{2} I_{1} = (3+2\sqrt{2}) I_{1} =5.83 I_{1}$

$I_{min} = I_{1} + I_{2} + 2I_{1}I_{2} = I_{1} +2 I_{1} -2\sqrt{2} I_{1} = (3-2\sqrt{2}) I_{1}=0.1716 I_{1}$

Initially when both slits have same intensity then bright fringe intensity is

$4I_{2} = 8 I_{1}$ and dark fringe is 0

So we see that bright fringe would be bright with reduced intensity and dark fringe would have some intensity now.

$\therefore$ Answer C) the dark fringes would be bright and bright fringes would be darker

The wave fronts of light wave traveling in vacuum are given by

$x+y+z=c$ . The angle made by the light ray with the x-axis is:

0%
0 $^{\circ}$
0%
45 $^{\circ}$
0%
90 $^{\circ}$
0%
$cos^{-1}\dfrac{1}{\sqrt{3}}$

Explanation

Normal to the wave front is
(1,1,1)
So, angle between normal and x-axis is angle between light ray and x-axis
So,

$(i+j+k).(i)=\sqrt{3}(1)cos\theta$

$cos\theta =\dfrac{1}{\sqrt{3}}$

$\theta =cos^{-1}(1/\sqrt{3})$

$I_{o}$ is the intensity of the principle maximum in the single slit diffraction pattern then, with doubling the slit width, the intensity becomes:

0%
$I_{o}$
0%
$I_{o}$ / 2
0%
$\frac{1}{2}$ $I_{o}$
0%
4 $I_{o}$

Explanation

Intensity remains

$I_o$ because intensity of principle maximum is maximum intensity and maximum intensity is independent of slit width.

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

$36\%$ by reducing its width, then the intensity of light at the same point will be :

0%
90
0%
89
0%
67
0%
81

Explanation

Maximum intensity

$I_{max}=I_1+I_2+2\sqrt{I_1I_2}$

Given :

$I_1 = I_2 = I$ and

$I_{max} =100$ units

$\therefore I_{max}=I+I+2\sqrt{II}$

$100 = 4I \Rightarrow I=25$

After reducing intensity of one of sources by

$36$ %, new intensity

$I' = \dfrac{64}{100}*25 =16$

New maximum intensity

$I'_{max}=I_1+I'+2\sqrt{I_1I'}$

$I'_{max}=25+16+2\sqrt{25\times 16}$

$=25+16+2\times 5\times 4$

$=41+40$

$=81 units.$

Two waves

$y_{1}=A_{1}sin (\omega t-\beta _{1})$ and

$y_{2}=A_{2}sin (\omega t-\beta _{2})$ superimpose to form a resultant wave whose amplitude is :

0%
$\sqrt{A_{1}^{2}+A^{2}_{2}+2A_{1}A_{2}cos(\beta _{1}-\beta _{2})}$
0%
$\sqrt{A_{1}^{2}+A^{2}_{2}+2A_{1}A_{2}sin(\beta _{1}-\beta _{2})}$
0%
$A_{1}+A_{2}$
0%
$|A_{1}+ A_{2}|$

Explanation

Resultant wave

$y=y_{1}+y_{2}$

$y=A_{1}sin(wt- \beta_{1})+A_{2}sin(wt- \beta_{2})$

$=A_{1}cos \beta_{1}sinwt-A_{1}coswt sin \beta_{1}+A_{2}sinwt cos \beta_{2}-A_{2}coswt sin\beta_{2}$

$=(A_{1}cos \beta_{1}+A_{2}cos \beta_{2})sinwt-(A_{1}sin \beta_{1}+A_{2}sin \beta_{2})coswt$

Amplitude of this wave is given by

$\sqrt{(A_{1}cos \beta_{1}+A_{2}cos \beta_{2})^{2}+(A_{1}sin \beta_{1}+A_{2}sin \beta_{2})^{2}}$

$=\sqrt{A_{1}^{2}cos^{2} \beta_{1}+A_{2}^{2}cos^{2} \beta_{2}+2A_{1}A_{2}cos \beta_{1}cos \beta_{2}+A_{1}^{2}sin^{2} \beta_{1}+A_{2}^{2}sin^{2} \beta_{2}+2A_{1}A_{2}sin \beta_{1}sin \beta_{2}}$

$=\sqrt{A_{1}^{2}+A_{2}^{2}+2A_{1}A_{2}cos( \beta_{1}- \beta_{2})}$

A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.0 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

We know

$b sin \theta =n\lambda$
b is the slit width ; n is 1 for

$1^{st}$ dark fringe

$sin \theta =\dfrac{1\times 600\times 10^{-9}m}{1\times 10^{-3}m}=6\times 10^{-4}$

$sin \theta =\dfrac{distance\ of\ 1^{st}\ dark\ fringe\ from \ center}{distance \ of \ screen}$
distance of

$1^{st}$ dark fringe

$=6\times 10^{-4}\times 2$ from center

$=1.2 mm$
As both are equidistant we have

$1.2 mm + 1.2 mm$
Distance between dark fringes

$=2.4mm$ .

In Young's double slit experiment the intensity of the maxima is

$I$ . If the width of each slit is doubled, the intensity of the maxima will be:

0%
$I/2$
0%
$2I$
0%
$4I$
0%
$I$

Explanation

$I=I_{max}=4I_{0}$
Now,

$I_{0}$ is increased to

$2I_{0}$
So,

$I_{max}=4(2I_{0})=8I_{0}=2I$
So, maximum intensity is

$2I$

A screen is placed 2m away from a single narrow slit. The slit width if the first minimum lies 5 mm on either side of the central maximum is:

(wave length

$=$ 5000A

$^{\circ}$ )

0%
$0.01cm$
0%
$0.02cm$
0%
$0.03cm$
0%
$0.04cm$

Explanation

$x=\dfrac{\lambda \times D}{\omega }$

$\omega =\dfrac{\lambda \times D}{x}$

$=\dfrac{5000\times 10^{-10}\times 2}{5\times 10^{-3}}$

$=2\times 10^{-4} m$

$=0.02 cm$

The diameter of the objective of a telescope is

$a$ , its magnifying power is

$m$ and wavelength of light

$\lambda$ . The resolving power of the telescope is :

0%
$\dfrac{(1.22\lambda )}{a}$
0%
$\dfrac{1.22a}{\lambda}$
0%
$\lambda (1.22a)$
0%
$\dfrac {a} {1.22\lambda}$

Explanation

Resolving power of telescope:

$R=\dfrac{1}{ \theta}=\dfrac{a}{1.22 \lambda}$

where

$\theta$ is angular resolution, a is diameter of the objective and

$\lambda$ is wavelength of light.

Yellow light is used in a single slit diffraction experiment with a slit of width

$0.6 mm$ . If yellow light is replaced by

$X-$ rays, then the observed pattern will reveal :

0%
more number of fringes
0%
less number of fringes
0%
no diffraction pattern
0%
that the central maximum is narrower

Explanation

As x-rays has more energy than yellow light it has less wavelength than yellow light.
The angular size of the central maxima is

$\theta =\dfrac{n \lambda }{a}$ where, n is integer

$(\neq 0)$

$\lambda$ is wavelength of light used a is slot width.
So, as '

$\lambda$ ' is decreased by using x-rays in place of yellow light the central maxima becomes narrower.

A person wants to see two pillars from a distance of 11 km, separately. The distance between the pillars must be approximately

0%
3.2m
0%
1m
0%
0.25 m
0%
0.5 m

Explanation

Resolving power of eye =

$1' = \dfrac {\pi} {60 \times 180} radians$ (approximately)

Let the separation between pillars

$= x m$

Distance

$= 11000 m$

$\dfrac {x} {11000} = \dfrac {\pi}{60 \times 180}$

$x = 3.2 m$ (approximately)

Answer. A)

$3.2m$

ASSERTION: Resolving power of telescope is more if the diameter of the objective lens is more.
REASON:Objective lens of large diameter collects more light.

0%
both A and R are correct and R is correct explanation of A
0%
A and R both are correct but R is not correct explanation of A
0%
A is true but R is false
0%
both A and R is false

Explanation

We have,

$RP=\dfrac { D }{ 1.22 \lambda}$

Hence

$R$ is correct but for larger resolution objects making small angle be distinguished or very close objects should be distinguished.

A parallel beam of monochromatic light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is :

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

Explanation

The phase difference between the rays coming from the two edges of the slit is given by the formula

$\dfrac{2\pi }{\lambda }dsin\theta$

where, d is the width of the slit.

the first minimum occurs when the following condition is satisfied

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

$\therefore \dfrac{2\pi }{\lambda }d\dfrac{\lambda }{d}$

$= 2\pi$

Match the following :

PART-A	PART-B
a) Polarisation	e) All types of waves
b) interference	f) longitudinal waves
c) diffraction	g) transverse waves
d) reflection	h) only with transverse waves
	i) stationary waves produced in stretched strings

0%
a $\rightarrow$ g ; b $\rightarrow$ e, f , g,i ; c $\rightarrow$ e, f , g ; d $\rightarrow$ e, f , g
0%
a $\rightarrow$ h, g; b $\rightarrow$ f , g ; c $\rightarrow$ g ; d $\rightarrow$ h
0%
a $\rightarrow$ e, f , g; b $\rightarrow$ g ; c $\rightarrow$ e, f , g ; d $\rightarrow$ g
0%
a $\rightarrow$ e; b $\rightarrow$ h,i ; c $\rightarrow$ g, h ; d $\rightarrow$ e

The limit of resolution of microscope, if the numerical aperture of microscope is 0.12, and the wavelength of light used is 600 nm, is

0%
0.3 $\mu$ m
0%
1.2 $\mu$ m
0%
2.5 $\mu$ m
0%
3 $\mu$ m

Explanation

For a microscope, the limit of resolution is given by,

$X = \dfrac{\lambda}{2A}$
where

$\lambda$ is the wavelength of light used, and A is the numerical aperture.
Hence, substituting the values,

$X = \dfrac{600}{2 \times 0.12}$ ,
which gives,

$X= 2.5 \mu m$

Light waves travel in vaccum along the y-axis. Then the wave front is:

0%
$y=$ constant
0%
$x=$ constant
0%
$z=$ constant
0%
$x+y+z =$ constant

Explanation

waves front are plane perpendicular to the direction of rays.
so, as light is traveling along y - axis, plane perpendicular to y - axis is the x-z plane with any constant value of y.
so, y

$=$ constant is the wave front plane

If accelerating potential increases from

$20\ KV$ to

$80\ KV$ in an electron microscope, its resolving power

$R$ would change to

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

Explanation

$\dfrac{1}{2}mv^{2}= eV$

$mv= \sqrt{2eVm}$

And

$\lambda = \dfrac{h}{mV}$

$\dfrac{\lambda _{0}}{\lambda _{1}}= \dfrac{\sqrt{2eV_{1}m}}{\sqrt{eV_{2}m}}$

$\dfrac{\lambda _{2}}{\lambda _{1}}= \dfrac{1}{2}$

$\therefore \lambda _{2}=\dfrac{\lambda _{1}}{2}$

$R\ \propto \dfrac{1}{\lambda}$

so

$R$ would change to

$2R$ .

The magnitude of magnifying power of an astronomical telescope is 5, the focal power of its eyepiece is 10 diopters. The focal power of its objective (in diopters) is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$M=5$

$P_{e}=10$

$\dfrac{1}{f_{e}}=10$

$f_{e}=\dfrac{1}{10}m=10cm$

$M=\dfrac{f_{0}}{f_{e}}=5$

$f_{0}=50 cm$

$P=2D$

A telescope has an objective lens of 10 cm diameter and is situated at a distance of

$1km$ for two objects. The minimum distance between these two objects, which can be resolved by the telesope, when the mean wavelength of light is 5000Å is of the order of

0%
5 cm
0%
0.5 mm
0%
5 m
0%
5 mm

Explanation

Resolution power

$= \dfrac {d\lambda}{D} = \dfrac {1000 \times 5000 \times 10^{-10}}{10 \times 10^{-2}} = 5mm$

To demonstrate the phenomenon of interference of sound, we need:

0%
two sources, which emit sound of exactly the same frequency
0%
two sources, which emit sound of exactly the same frequency and have a definite phase relationship
0%
two sources, which emit sound of exactly the same frequency and have a varying phase relationship
0%
two sources, which emit sound of exactly the same wavelength

Explanation

For demonstration of interference of sound we need two coherent source, of same frequency with constant phase difference

$P_1=P_{01} sin (Kx-\omega t)$

$P_2=P_{02} sin (k(x+\Delta x)-\omega t)$

$=P_{02} sin (Kx-\omega t+\delta )$

$\delta =K\Delta x=\dfrac{2\pi \Delta x}{\lambda }$

Two coherent sources of intensity ratio $\beta$ interfere. Then the value of $\displaystyle \left( {\frac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}}} \right)$ is:

0%
$\dfrac{{1 + \beta }}{{\sqrt \beta }}$
0%
$\sqrt {\left( {\dfrac{{1 + \beta }}{\beta }} \right)}$
0%
$\dfrac{{1 + \beta }}{{2\sqrt \beta }}$
0%
$\dfrac{{2\sqrt \beta }}{{1 + \beta }}$

Explanation

$\displaystyle \frac{I_2}{I_1}=\beta$

$\displaystyle \frac{I_{min}}{I_{max}}=\frac{\beta - 2 \sqrt{\beta}+1}{\beta + 2 \sqrt{\beta +1}}$

$\frac{A_2}{A_1} =\sqrt{\beta}$

$\displaystyle \frac{A_2 -A_1}{A_2 + A_1}=\frac{\sqrt{\beta}-1}{\sqrt{\beta}+1}$

$\displaystyle \frac{I_{max}- I_{min}}{I_{max}+ I_{min}}=\frac{(2)}{(2)} \left [ \frac{2 \sqrt{\beta}}{\beta +1}\right ]$

The resolution limit of the eye is

$1'$ . At a distance

$x\ km$ from the eye, two persons stand with a lateral separation of 3 meter. For the two persons to be just resolved by the naked eye,

$x$ should be:

0%
$10 km$
0%
$15 km$
0%
$20 km$
0%
$30 km$

Explanation

Distance =

$x km = 1000x m$

Resolution limit

$= 1'$

Lateral separation

$= 3$ meter

Angle on eye in minutes for 2 person,

$x$ km away separated by

$3$ meter

$=$

$\dfrac{3}{1000x} \times \dfrac{180}{ \pi} \times 60$

So,

$1 =$

$\dfrac{3}{1000 x} \times \dfrac{180}{\pi} \times 60$

$\Rightarrow x = 10.3$ Km ( approximately

$10$ Km)

Answer. A)

$10$ Km

An astronomical telescope has a large aperture to

0%
reduce spherical aberration
0%
have high resolution
0%
increase span of observation
0%
have low dispersion

Two point white dots are

$1\ mm$ apart on a black paper. They are viewed by eye of pupil diameter

$3\ mm$ . Approximately, what is the maximum distance at which these dots can be resolved by the eye? (Take wavelength of light

$= 500\ nm$ )

0%
6 m
0%
3 m
0%
5 m
0%
1 m

Explanation

$R.P. = \dfrac{2 \mu \sin \theta}{1.22 \lambda}$

$\dfrac{1}{10^{-3}} = \dfrac{2 \sin \theta}{1.22 \times 5 \times 10^{-7}}$

$2 \theta = 6 \times 10^{-4}$

$\dfrac{d}{D} = 6 \times 10^{-4}$

$\therefore D = \dfrac{3 \times 10^{-3}}{6 \times 10^{-4}} = 5 m$

In double slit experiment, for light of which colour, the fringe width will be minimum?

0%
Violet
0%
Red
0%
Green
0%
Yellow

Explanation

$\beta \propto \lambda.$

$\therefore \lambda_v =$ minimum.

Correct answer is violet, as it has minimum wavelength . Fringe width is directly proportional to the wavelength. So, purple or violet has the minimum fringe width whereas red has maximum fringe width.

A telescope has an objective lens of 10 cm diameter and is situated at a distance of one kilometer from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is

$5000 \dot{A}$ , is of the order of :

0%
5 m
0%
5 mm
0%
5 cm
0%
0.5 m

Explanation

We know,

Resolving Power =

$\theta = \dfrac{d} {D}$ but

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

$\left (Note : exact\ relation, \theta = \dfrac{1.22} {a}\lambda \right )$

$\dfrac{d} {D} = \dfrac{\lambda} {a} \Rightarrow = d = \dfrac{\lambda D} {a}$

$\Rightarrow d = \dfrac{5000 \times 10^{-10} \times 10^{3}} {10 \times 10^{-2}} = 5 mm$

Identify the correct statement from the following :

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Wave nature of light was proposed by Huygens.
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The direction of light ray and its wave front are opposite.
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Huygen's wave theory could not explain phenomenon of reflection.
0%
A monochromatic ray of light after passing through the prism should create a spectrum of seven colours.

The wave front is a surface in which:

0%
All points are in the same phase
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There is a pair of points in opposite phase
0%
There is a pair of points with phase difference $(\cfrac {\pi}{2})$
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There is no relation between the phase

The phenomenon of interference is shown by

0%
Longitudinal mechanical waves only
0%
Transverse mechanical waves only
0%
Non-mechanical transverse waves only
0%
All the above types of waves

Explanation

Correct option is : (D) All kind of waves

$\bullet$ Interference occurs in all kinds of waves, pattern depending upon the phases of the interfering waves.

$\bullet$ Interference in longitudinal waves occurs in sound waves.

$\bullet$ Interference in transverse mechanical waves occurs in water ripples.

$\bullet$ Interference in transverse non mechanical waves occurs in light.

CBSE Questions for Class 12 Medical Physics Wave Optics Quiz 4 - 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 4 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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