3 Marks Question

Question 513 Marks

Find the ratio of intensities at two points on a screen in Young’s double slit experiment when waves from the two slits have a path difference of (i) 0 and (ii) $\frac{\lambda}{4}.$

Answer

Intesity $\text{I}=\text{a}^2_1+\text{a}^2_2+2\text{a}_1\text{a}_2\cos\varphi$
Let $\text{a}_1=\text{a}_2=\text{a}$ say, then
$\text{I}=\text{a}^2+\text{a}^2+2\text{a}^2\cos\varphi=2\text{a}^2(1+\cos\varphi)$
$\frac{\text{I}_1}{\text{I}_2}=\frac{1+\cos\phi_1}{1+\cos\phi_2}$
When path difference is 0, phase difference $\varphi_1=0$
When path difference $\frac{\lambda}{4},$ is phase difference $\phi_2=\frac{2\pi}{\lambda}\times\frac{\lambda}{4}=\frac{\pi}{2} $
$\therefore\frac{\text{I}_1}{\text{I}_2}\frac{1+\cos\phi_1}{1+\cos\frac{\pi}{2}}=\frac{1+1}{1}=\frac{2}{1}.$

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Question 523 Marks

Sketch a graph showing the variation of fringe width versus the distance of the screen from the plane of the slits (keeping other parameters same) in Young’s double slit experiment. What information can one obtain from the slope of this graph?

Answer

We know that the fringe width is given by:$\beta=\lambda\frac{\text{D}}{\text{d}}$
$\Rightarrow \beta=\frac{\lambda}{\text{d}}\text{D}$
The graph between β and D is shown alongside
The slope of graph $=\frac{\lambda}{\text{d}}$

Knowing d, the wavelength of light used can be calculated to be
$\lambda= \text{Slope of graph}\times \text{d.}$

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Question 533 Marks

In Young’s double slit experiment, explain with reason in each case, how the interference pattern changes, when :

Width of the slits is doubled.
Separation between the slits is increased.
Screen is moved away from the plane of slits.

Answer

The fringe width $\beta= \frac{\text{D}\lambda}{\text{d}}.$
When the width of the slit is doubled; the intensity of interfering waves becomes four times, intensity of maxima becomes 16 times i.e., fringes become brighter.
When separation between the slits is increased the fringe width decreases, i.e., fringes come closer.
$\beta \alpha\text {D }$ when screen is moved away from the plane of the slits, the fringe width increases, i.e., fringes become farther away.

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Question 543 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 553 Marks

The index of refraction of fused quartz is 1.472 for light of wavelength 400nm and is 1.452 for light of wavelength 760nm. Find the speeds of light of these wavelengths in fused quartz.

Answer

We know that, $\frac{\mu_2}{\mu_1}=\frac{\text{v}_1}{\text{v}_2}$So, $\frac{1472}{1}=\frac{3\times10^8}{\text{v}_{400}}\Rightarrow\text{v}_{400}=2.04\times10^8\text{m/sec}.$
[because, for air, $\mu=1\ \text{and v}=3\times10^8\text{m/s}]$
Again, $\frac{1452}{1}=\frac{3\times10^8}{\text{v}_{760}}\Rightarrow\text{v}_{760}=2.07\times10^8\text{m/sec}.$

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Physics 3 Marks Question