MCQ Questions for Class 12 Medical Physics Wave Optics Quiz 7

White light is made of-

0%
2 colours
0%
4 colours
0%
7 colours
0%
3 colours

In case of linearly polarised light, the magnitude of the electric field vector.

0%
Does not change with time
0%
Varies periodically with time
0%
Increases and decreases linearly with time
0%
Is parallel to the direction of propagation

doctor prescribes spectacles to a patient with a combination of a convex lens of focal length 40 em, and concave lens of focal length 25 em then the power of spectacles will be.

0%
-6.5 D
0%
1.5 D
0%
-1.5 D
0%
-8.5 D

Explanation

For combination of lenses, power

$P=P_1+P_2=\frac {100}{40}-\frac{100}{25}=-1.5 D$

Assertion: Standard optical diffraction grating scan not be used for discriminating between X-ray wavelengths.
Reason: The grating spacing is not of the order of X-ray wavelengths.

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If both assertion and reason are true but the reason is the correct explanation of assertion
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If both assertion and reason are true but the reason is not the correct explanation of assertion
0%
If assertion is true but reason is false
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If both the assertion and reason are false
0%
If reason is true but assertion is false

Assertion: The resolving power of a telescope is more if the diameter of the objective lens is more.
Reason: Objective lens of large diameter collects more light.

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Both assertion and reason are true but the reason is the correct explanation of assertion
0%
Both assertion and reason are true but the reason is not the correct explanation of assertion
0%
Assertion is true but reason is false
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Both the assertion and reason are false
0%
Reason is true but assertion is false

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 wave length of light =500 nm]

0%
10 m
0%
5 m
0%
15 m
0%
None of these

Explanation

We have,

$\dfrac{y}{D}\ge 1.22 \dfrac{\lambda}{d}$

$\Rightarrow D \le \dfrac{yd}{(1.22)\lambda} = \dfrac{10^{-3}\times 3\times 10^{-3}}{(1.22)\times 5\times 10^{-7}}$

$=\dfrac{30}{6.1}\approx 5m$

$\therefore D_{max} = 5m$

Wavelength of light used in an optical instrument are

$\lambda _1 = 4000 \mathring { A }$ and

$\lambda _2 = 5000 \mathring { A}$ then ratio of their respective resolving powers(corresponding to

$\lambda _1$ and

$\lambda _2$ ) is

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16:25
0%
9:1
0%
4:5
0%
5:4

Explanation

We know,

$Resolving\ Power\ \propto \dfrac{1}{\lambda}$

Given,

$\lambda_1 = 4000 \mathring{A}$

$\lambda_2 = 5000 \mathring{A}$

$\Rightarrow \dfrac{RP_{\lambda_1}}{RP_{\lambda_2}} = \dfrac{ \lambda_2}{ \lambda_1}$

$\Rightarrow \dfrac{RP_{\lambda_1}}{RP_{\lambda_2}} = \dfrac{5000}{4000} = \dfrac{5}{4}$

$\Rightarrow RP_{\lambda_1} : RP_{\lambda_2} = 5:4$

Hence, the correct answer is OPTION D.

A: The corpuscular theory fails in explaining the velocities of light in air and water.
B: According to corpuscular theory, the light should travel faster in a 'denser medium than in a rarer medium.

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If both A and B are true but the B is the correct explanation of A
0%
If both A and B are true but the B is not the correct explanation of A
0%
If A is true but B is false
0%
If both the A and B are false
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If B is true but A is false

Assertion: Diffraction takes place for all types of waves mechanical or non-mechanical, transverse or longitudinal.
Reason: Diffraction's effect are perceptible only if wavelength of wave is comparable to dimensions of diffracting device.

0%
If both Assertion 'and Reason are correct and Reason is the correct explanation ofAssertion
0%
If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
0%
If Assertion is correct but Reason is incorrect
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If Assertion is incorrect but Reason is correct

statement 1: When a tiny circular obstacle is placed in the path of light from some distance, a bright spot is seen at the center of the shadow of the obstacle.
statement 2: Destructive interference occurs at the center of the shadow.

0%
If both statement 1 and statement 2 are true but statement 2 is the correct explanation of statement 1
0%
If both statement 1 and statement 2 are true but the reason is not the correct explanation of statement 1
0%
If statement 1 is true but statement 2 is false
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If both the statement 1 and statement 2 are false
0%
If statement 2 is true but statement 1 is false

Assertion: The clouds in sky generally appear to be whitish.
Reason: Diffraction due to clouds is efficient in equal measure at all wavelengths.

0%
If both assertion and reason are true but the reason is the correct explanation of assertion
0%
If both assertion and reason are true but the reason is not the correct explanation of assertion
0%
If assertion is true but reason is false
0%
If both the assertion and reason are false
0%
If reason is true but assertion is false

Assertion: Radio waves can'be polarised.
Reason: Sound waves in air are longitudinal in nature.

0%
If both assertion and reason are true but the reason is the correct explanation of assertion
0%
If both assertion and reason are true but the reason is not the correct explanation of assertion
0%
If assertion is true but reason is false
0%
If both the assertion and reason are false
0%
If reason is true but assertion is false

Huygen's wave theory allows us to know

0%
The wavelength of the wave
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The velocity of the wave
0%
The amplitude of the wave
0%
The propagation of the wavefront

Light travels faster in air than that in glass. This is accordance with

0%
wave theory of light
0%
corpuscular theory of light
0%
neither (a) nor (b)
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Both (a) and (b)

In Young's double slit experiment if the slit width is in the ratio

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

0%
$1:1$
0%
$1:9$
0%
$1:4$
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$1:3$

Explanation

Amplitude of light from a slit of width d

$\propto \sqrt{d}$

Hence

$\dfrac{A_1}{A_2}=\sqrt{\dfrac{1}{9}} =\dfrac{1}{3}$

$\implies A_2=3A_1$

Thus the ratio of intensity at minima to that at maxima

$=\dfrac{(A_2-A_1)^2}{(A_2+A_1)^2}$

$=\dfrac{2^2}{4^2}=\dfrac{1}{4}$

Four light sources produce the following four waves :
(I) $y_1 \, = \, a' \, \sin \,(\omega t \, + \,\phi_1)$
(ii) $y_2 \, = \, a' \, \sin \,(2 \omega t)$
(iii) $y_3 \, = \, a' \, \sin \,(\omega t \, + \, \phi_2)$
(iv) $y_4 \, = \, a' \, \sin \,(3 \omega t \, + \, \phi)$
super position of which two waves give rise to interference ?

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(i) and (ii)
0%
(ii) and (iii)
0%
(i) and (iii)
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(iii) and (iv)

Explanation

Interference phenomenon takes place between two waves which have equal frequency and propagate in same direction.
Hence ,

$y_1 \, = \, a \, \sin \,( \omega t \,+\, \phi_1)$

$y_3 \, =\, a' \, \sin \,( \omega t \,+\, \phi_2)$
will give rise to interference as the two waves have same frequency

$\omega.$

Which of the following generates a plane wavefront?

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$\alpha-rays$
0%
$\beta-rays$
0%
$\gamma-rays$
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None of these

A light wave is incident normally over a slit of width

$24\times {10}^{-5}cm$ . The angular position of second dark fringe from the central maxima is

${30}^{o}$ . What is wavelength of light?

0%
$6000\mathring { A }$
0%
$5000\mathring { A }$
0%
$3000\mathring { A }$
0%
$1500\mathring { A }$

Explanation

For second dark fringe,

$d\sin{\theta}=2\lambda$

$\Rightarrow$

$24\times {10}^{-5}\times {10}^{-2}\times \sin{{30}^{o}}=2\lambda$

$\Rightarrow$

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

$=6000\mathring { A }$

In a single slit diffraction pattern fringes are of:

0%
unequal width
0%
equal width
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equal width and equal intensity
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unequal width and unequal intensity

Considering a beam of light is incident on a rectangular opening in the front of a box, as shown in the side view above. The back of the box is open. The light is incident on a screen after passing through the box. The following devices may be in the box, positioned as shown.
Which device could produce a diffraction pattern consisting of a central bright fringe with parallel secondary fringes that decrease in intensity with increasing distance from the center of the screen?

Light of wavelength

$\displaystyle \lambda$ 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%
$\displaystyle \lambda \approx \frac { { d }^{ 2 } }{ 4D }$
0%
$\displaystyle d\approx \frac { { \lambda }^{ 2 } }{ D }$
0%
$\displaystyle \lambda \approx \frac { D }{ 4 }$
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$\displaystyle \sqrt { \lambda } =\frac { d }{ 4D }$

Explanation

Rings are formed when constructive interference occurs between the light waves from two slits. Condition for this to happen is:

$\lambda \approx \dfrac{d^2}{4D}$

The phenomenon of rotation of plane polarized light is called:

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Kerr effect
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Double refraction
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Optical activity
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Dichroism

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

$0.6\ mm$ . If yellow light is replaced by X-rays, then the pattern will reveal:

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

Explanation

Wavelength of X-Rays ranges from

$0.1$ to

$10\ nm$ and yellow light is in the range

$570$ to

$590\ nm$ .

Hence,

$\lambda_{X-ray}<\lambda_{yellow}$

Fringes are formed at

$x_n=\dfrac{n\lambda D}{d}$

Hence, fringes of X-Rays are formed close to each other and more number of fringes.

A point source of light is placed at origin, in air. The equation of wave front of the wave at time

$t$ , emitted by source at

$t=0$ , is (take refractive index of air as

$1$ and speed of light to be

$e$ )

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$x+y+z=et$
0%
$x^{2}+y^{2}+z^{2}=t^{2}$
0%
$xy+yz+zx=e^{2}t^{2}$
0%
$x^{2}+y^{2}+z^{2}=e^{2}t^{2}$

In Young's double slit experiment a minima is observed when path difference between the interfering beam is

0%
$\lambda$
0%
$1.5\lambda$
0%
$2\lambda$
0%
$2.25\lambda$

Explanation

The condition for minimum intensity, when the path difference is a multiple of half wavelengths, i.e.,

$d\sin\theta = \left(n+\dfrac 12 \right) \lambda$

So for the minima, n=1,

$\implies d \sin\theta=1.5\lambda$

Which of the following statement is false:

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Sound and light wave exhibit interference
0%
Sound and light wave exhibit diffraction
0%
Light wave exhibits polarization while sound wave does not
0%
Sound wave exhibits polarization while light wave does not

How to increase the widths of the fringes in the diffraction pattern that appears on the screen in a double-slit interference experiment with light?

0%
Use light of a shorter wavelength
0%
Move the screen closer to the slits
0%
Move the slits closer together
0%
Use light with a lower wave speed
0%
Increase the intensity of the light

Explanation

We know the fringe width

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

where

$D=$ distance between slits and screen

$\lambda=$ wavelength of light used

$d=$ distance between two slits

by above formula

$\beta \propto \dfrac{1}{d}$

therefore fringe width increases when distance between two slits will decrease or we can say that we move the slits closer together.

A plane wave of wavelength

$6250$

$\displaystyle \overset { \circ }{ A }$ is incident normally on a slit of width

$\displaystyle 2\times { 10 }^{ -2 }cm$ . The width of the principle maximum of diffraction pattern on a screen at a distance of

$50 \ cm$ will be:

0%
$\displaystyle 312\times { 10 }^{ -3 }cm$
0%
$\displaystyle 312.5\times { 10 }^{ -4 }cm$
0%
$\displaystyle 312cm$
0%
$\displaystyle 312.5\times { 10 }^{ -5 }cm$

Explanation

For single slit diffraction,

$asin\theta=\lambda$ for first minima

$\implies a\dfrac{y}{D}=\lambda$

$\implies y=\dfrac{D\lambda}{a}$

$=\dfrac{0.5\times 6250\times10^{-10}}{2\times 10^{-4}}$

$=1.56\times 10^{-3}m$

Thus, the width of the principle maxima=

$2y=3.12\times 10^{-3}m$

$=312\times 10^{-3}cm$

Which combination of wavelength and lines/meter will cause the first bright fringe to be closest to straight ahead on a screen in front of a diffraction grating?

0%
wavelength - lines/meter, $400 nm - 100,000$
0%
wavelength - lines/meter, $500 nm - 300,000$
0%
wavelength - lines/meter, $600 nm - 100,000$
0%
wavelength - lines/meter, $700 nm - 200,000$
0%
wavelength - lines/meter, $500 nm - 100,000$

Explanation

The relationship between the wavelength of light

$\lambda$ and angle

$\theta$ is

$d\sin\theta =n\lambda$ , where d is the spacing of adjacent lines on the grating, and n is the order number.

For first bright fringe to be closest to straight ahead on a screen in front of a diffraction grating,

$\dfrac \lambda d=\lambda N$ should be minimum.

Here only wavelength 400nm and 100,000 lines/meter satisfy this condition.

Huygen's originally thought that for the propagation of light waves wavefront is required.

0%
True
0%
False

Red light has a longer wavelength than green light which has a longer wavelength than blue light which has a longer wavelength than violet light.
When white light passes through a diffraction grating, which order is "bent" by diffraction the most?

0%
Red
0%
Green
0%
Blue
0%
Violet
0%
All colors "bend" the same

Explanation

Using

$b sin\theta = n\lambda$

$\therefore$

$\theta = sin^{-1}\bigg( \dfrac{n\lambda}{b} \bigg)$

$\implies \theta \propto \lambda$

Given :

$\lambda_{red} > \lambda_{green} >\lambda_{blue}>\lambda_{violet}$

$\implies$

$\theta_{red} > \theta_{green} > \theta_{blue} >\theta_{violet}$

Hence the red light bends the most.

A teacher stands in front of the door of a classroom, and only one row of students sitting at their desks can see the teacher. The teacher speaks from the same location, and all the students in the classroom (five rows) are able to hear.
Since both the light waves and the sound waves from the teacher pass through the door, why can all students hear the teacher but not see the teacher?

0%
The sound waves are refracted when they pass through the door, but the light waves travel too fast to be refracted when they pass through the door
0%
The particle nature of light prevents light from "bending" as it passes through the door
0%
Noticeable diffraction of waves can occur when the width openings are close to the wavelengths of the waves being diffracted. Sound waves are noticeably diffracted as they pass through the door, but light wavelengths are way too short to be noticeably diffracted
0%
Sound waves have the property of spreading out when they pass through opening; light waves never do
0%
Longitudinal waves (like sound waves) are always spreading out in all direction; transverse waves (like light waves) travel only in straight lines

Which combination of wavelength and slit width gives the widest central bright spot for single-slit diffraction?

0%
Wavelength- slit width, $600 nm - 600 nm$
0%
Wavelength- slit width, $600 nm - 300 nm$
0%
Wavelength- slit width, $500 nm - 300 nm$
0%
Wavelength- slit width, $400 nm - 800 nm$
0%
Wavelength- slit width, $400 nm - 400 nm$

Explanation

Using

$bsin\theta = n\lambda$

$\implies \theta \propto \dfrac{\lambda}{b} = k \dfrac{\lambda}{b}$ where

$k =$ constant

(A) :

$\theta_A =k \times \dfrac{600}{600} = k$

Similarly,

$\theta_B = 2k$ ,

$\theta_C = 1.67k$ ,

$\theta_D = 0.5k$ ,

$\theta_E= k$

Thus data given in option B will make the widest central bright spot.

A physicist shines coherent light through an object, A , which produces a pattern of concentric rings on a screen, B. A is most likely:

0%
A polarization filter
0%
A single-slit
0%
A multiple-slit diffraction grating
0%
A prism
0%
A sheet with a pinhole

The solar glare of sunlight bouncing off water or snow can be a real problem for drivers. The reflecting sunlight is horizontally polarized, meaning that the light waves oscillate at an angle of

$90^o$ to a normal line drawn perpendicular to the Earth. At what angle relative to this normal line should sunglasses be polarized if they are to be effective against solar glare?

0%
$0^o$
0%
$30^o$
0%
$45^o$
0%
$60^o$
0%
$90^o$

Explanation

The idea behind polarized sunglasses is to eliminate the glare. If the solar glare is all at a

$90^o$ angle to the normal line, sunglasses polarized at a

$0^o$ angle to this normal will not allow any of the glare to pass. Most other light is not polarized, so it will still be possible to see the road and other cars, but the distracting glare will cease to be a problem.

Find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of a slit of width

$12\times { 10 }^{ 5 }cm$ when the slit is illuminated by monochromatic light of wavelength 6000

$\mathring { A } \quad$ .

0%
${ 40 }^{ \circ }$
0%
${ 45 }^{ \circ }$
0%
${ 30 }^{ \circ }$
0%
${ 60 }^{ \circ }$

Indentify the correct statement from the following

0%
Wave nature of light was proposed by Huygen.
0%
The direction of light ray and its wave front are opposite.
0%
Huygen's wave theory could not explain phenomenon of reflection.
0%
A monochromatic ray of light after passing through the prism should be made of one colour only.

A diffraction pattern is obtained using a beam of red light. Which one among the following will be outcome if the redlight is replaced by blue light?

0%
Bands disappear
0%
Diffraction pattern becomes broader and further apart
0%
Diffraction pattern becomes narrower and crowded together
0%
No change

The optical instrument which is used in every cricket match is.:

0%
Simple microscope
0%
Compound microscope
0%
Astronomical telescope
0%
Binocular

The wavefront is a surface in which

0%
all points are in the same phase
0%
there is a pair of points in opposite phase
0%
there is a pair of points with phase difference $(\dfrac{\pi}{2})$
0%
there is no relation between the phases

Diameter of the objective of a telescope is 200 cm. What is the resolving power of a telescope? Take wavelength of light = 5000

$\mathring{A}$ .

0%
$6.56 \times 10^6$
0%
$3.28 \times 10^5$
0%
$1 \times 10^6$
0%
$3.28 \times 10^6$

Explanation

Given :

$D = 200$ cm

$=2$ m

$\lambda = 5000$

$oA = 5\times 10^{-7}$ m

Resolving power of telescope

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

$\therefore$

$RP = \dfrac{2}{1.22 \times 5\times 10^{-7}} = 3.28\times 10^6$

With the help of a telescope that has an objective of diameter

$200\ cm$ , it is proved that light of wavelengths of the order of

$6400\overset {\circ}A$ coming from a star can be easily resolved. Then the limit of resolution is

0%
$39\times 10^{-8}deg$
0%
$39\times 10^{-8} rad$
0%
$19.5\times 10^{-8}rad$
0%
$19.5\times 10^{-8}deg$

Explanation

Limit of resolution

$=\theta = \dfrac{1.22\times \text{wavelength of light}}{\text{diameter of objective}}=\dfrac {1.22\lambda}{d} = \dfrac {1.22\times 6400\times 10^{-10}}{200\times 10^{-12}} = 39\times 10^{-8}rad$

Fill in the blank.

The phenomenon of polarization shows that light has ________ nature.

0%
Particle
0%
Transverse
0%
Longitudinal
0%
Dual

In the diffraction pattern due to a single slit of width '

$d$ ' with incident light of wavelength '

$\lambda$ ' , at an angle of diffraction '

$\theta$ ' , the condition for first minimum is ____________.

0%
$\lambda \sin { \theta } =d$
0%
$d\cos { \theta } =\lambda$
0%
$d\sin { \theta } =\lambda$
0%
$\lambda \cos { \theta } =d$

Explanation

The

$n^{th}$ order minima of a diffraction pattern is obtained as

$d sin\theta=n\lambda$

Hence the first minima is given as

$d sin\theta=(1)\lambda$

$\implies d sin\theta=\lambda$

A light of wavelength

$400\overset{o}{A}$ after travelling a distance of

$2\mu m$ produces a phase change of:

0%
Zero
0%
$3\pi$
0%
$\displaystyle\frac{\pi}{2}$
0%
$\displaystyle\frac{\pi}{3}$

Explanation

The number of wavelengths the wave travels is

$=\dfrac{l}{\lambda}$

$=\dfrac{2\times 10^{-6}}{400\times 10^{-10}}$

$=50$ which is an integer.

Hence, the point is at a distance which is integral multiple of wavelength.

Thus the phase difference between the points is zero.

$i$ is the angle of incidence, the angle between the incident wave front and the normal to the reflecting surface is

0%
$i$
0%
${ 90 }^{ o }-i$
0%
${ 90 }^{ o }+i$
0%
$i-{ 90 }^{ o }$

Explanation

A case of wavefront incident on the reflecting surface is shown in the figure above. The line AE represents incident wavefront at time of incidence. The angle between the wavefront and normal to the reflecting surface here is

$90^{\circ}-i$ .

When two waves of almost equal frequency

$n_1$ and

$n_2$ are produced simultaneously, then the times interval between successive maxima is

0%
$\dfrac{1}{n_1+n_2}$
0%
$\dfrac{1}{n_1}+\dfrac{1}{n_2}$
0%
$\dfrac{1}{n_1}-\dfrac{1}{n_2}$
0%
$\dfrac{1}{n_1-n_2}$

Explanation

The intensity of the sound is maximum in all these cases,

$2 \pi(\dfrac{n_1-n_2}{2})\times t=n\pi$

$t=0$ ,

$\dfrac{1}{n_1-n_2}$ ,

$\dfrac{2}{n_1-n_2}....$

Time interval between two successive maxima = time interval between two successive beats

$= \dfrac{1}{n_1-n_2}$

The wave theory of Light was proposed by ...............

0%
Newton
0%
Huygens
0%
Frenel
0%
Yung

In a plane transmission grating, the width of a ruling is

$12000\overset{o}{A}$ and the width of a slit is

$8000\overset{o}{A}$ , the grating element is:

0%
$20\mu m$
0%
$2\mu m$
0%
$200\mu m$
0%
$10\mu m$

Explanation

Grating element is the sum of width of ruling and width of slit.

Grating element is

$12000A^{0} + 8000A^{0}=20000A^{0}$ or

$2\mu m$

A light of wavelength

$6000\mathring { A }$ is incident normally on a grating

$0.005 m$ wide with

$2500$ lines. Then the maximum order is :

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

Explanation

Length of grating is

$0.005/2500=2\times10^{-6}m$

Now,

$m\lambda=e$ (for maximum order)

$m\times6000A^0 = 2\times10^{-6}$

$m=3$

CBSE Questions for Class 12 Medical Physics Wave Optics Quiz 7 - 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 7 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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