3 Marks Question

Question 513 Marks

A narrow slit S transmitting light of wavelength $\lambda$ is placed a distance d above a large plane mirror as shown in figure (17-E1). The light coming directly from the slit and that coming after the reflection interfere at a screen $\sum$ placed at a distance D from the slit.

What will be the intensity at a point just above the mirror, i.e., just above O?
At what distance from 0 does the first maximum occur?

Answer

Since, there is a phase difference of $\pi$ between direct light and reflecting light, the intensity just above the mirror will be zero.
Here, 2d = equivalent slit separation

D = Distance between slit and screen.

We know for bright fringe, $\Delta\text{x}=\frac{\text{y}\times2\text{d}}{\text{D}}=\text{n}\lambda$

But as there is a phase reversal of $\frac{\lambda}{2}.$

$\Rightarrow\frac{\text{y}\times2\text{d}}{\text{D}}+\frac{\lambda}{2}=\text{n}\lambda$

$\Rightarrow\frac{\text{y}\times2\text{d}}{\text{D}}=\text{n}\lambda-\frac{\lambda}{2}\Rightarrow\text{y}=\frac{\lambda\text{D}}{4\text{d}}$

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Question 523 Marks

What is the shape of the wavefront on earth for sunlight?

Answer

The sun is at very large distance from the earth. Assuming sun as spherical, it can be considered as point source situated at infinity. We can treat it like a point object as seen from the surface of earth.
Because of large distance, the radius of wavefront can be considered as Large (infinity) and hence, wavefront is almost plane.

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Question 533 Marks

A parallel beam of light of wavelength 600nm is incident normally on a slit of width ‘a’. If the distance between the slit and the screen is 0.8 and the distance of 2nd order maximum from the centre of the screen is 1.5mm, calculate the width of the slit.

Answer

Given: $\lambda=600\text{nm}=600\times10^{-9}\text{m}=6.0\times10^{-7}\text{m},\text{D}=0.8\text{m,}$
$\text{y}_2=1.5\text{mm}=1.5\times10^{-3}\text{m},\text{n}=2,\text{a=?}$
Position of $n^{th}$ maximum in diffraction of a single slit
$\text{y}_n=\text{(n}+\frac{1}{2})\frac{\lambda\text{D}}{\text{a}}\Rightarrow\text{a}=(\text{n}+\frac{1}{2})\frac{\lambda\text{D}}{\text{y}_\text{n}}$
Substituting given value $\text{a}=(2+\frac{1}{2})\frac{6.0\times10^{-7}\times0.8}{1.5\times10^{-3}}$
$\frac{5}2\times4.0\times0.8\times10^{-4}\text{m}=0.8\times10^{-3}=0.8\text{mm}.$

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Question 543 Marks

Two coherent waves of equal amplitude produce interference pattern in Young’s double slit experiment. What is the ratio of intensity at a point where phase difference is $\frac{\pi}{2}$ to intensity at centre?

Answer

$\text{I}_\frac{\pi}{2}=\text{a}^2_1+\text{a}^2_2+\text{2a}_1\text{a}_2\cos\varphi$
$=\text{a}^2+\text{a}^2+2\text{a}^2\cos\frac{\pi}{2}=2\text{a}^2$
$\text{I}_\max=(\text{a}_2+\text{a}_2)^2=(\text{a}+\text{a})^2=4\text{a}^2$
$\therefore\frac{\text{I}_\frac{\pi}{2}}{\text{I}_\max}=\frac{2\text{a}^2}{\text{4a}^2}=\frac{1}{2}.$

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Question 553 Marks

Two narrow slits emitting iight in phase are separated by a distance of 1.0cm. The wavelength of the light is $5.0 \times 10^{-7}m.$ The interference pattern is observed on a screen placed at a distance of 1.0m.

Find the separation between the consecutive maxima. Can you expect to distinguish between these maxima?
Find the separation between the sources which will give a separation of 1.0mm between the consecutive maxima.

Answer

Given that, $d = 1cm = 10^{-2}m, \lambda=5\times10^{-7}\text{m}$ and D = 1m

Separation between two consecutive maxima is equal to fringe width.

So, $\beta=\frac{\lambda\text{D}}{\text{d}}=\frac{5\times10^{-7}\times1}{10^{-2}}\text{m}=5\times10^{-5}\text{m}=0.05\text{mm}.$

When, $\lambda=1\text{mm}=10^{-3}\text{m}$

$10^{-3}\text{m}=\frac{5\times10^{-7}\times1}{\text{D}}\Rightarrow\text{D}=5\times10^{-4}\text{m}=0.50\text{mm}.$

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Physics 3 Marks Question