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Physics · 2024 · 5 marks
CBSE 2024 · Region 2 · Set 1 · Q33
(a) (i) Give any two differences between the interference pattern obtained in Young's double-slit experiment and a diffraction pattern due to a…
(i)
Give any two differences between the interference pattern obtained in Young's double-slit experiment and a diffraction pattern due to a single slit.
(ii)
Draw an intensity distribution graph in case of a double-slit interference pattern.
(iii)
In Young's double-slit experiment using monochromatic light of wavelength $\displaystyle \lambda$, the intensity of light at a point on the screen, where path difference is $\displaystyle \lambda$, is K units. Find the intensity of light at a point on the screen where the path difference is $\displaystyle \frac{\lambda}{6}$.
(i)
Draw a labelled ray diagram of a compound microscope showing image formation at least distance of distinct vision. Derive an expression for its magnifying power.
(ii)
A telescope consists of two lenses of focal length $\displaystyle 100$ cm and $\displaystyle 5$ cm . Find the magnifying power when the final image is formed at infinity.
Marking-scheme solution
(i)
Interference vs Diffraction:
1. Bands are equally spaced | Bands are not equally spaced.
2. Intensity of bright bands is the same | Intensity of maxima decreases on either side of central maximum.
3. First maximum is at an angle $\displaystyle \lambda / a$ | First minimum is at an angle $\displaystyle \lambda / a$
(iii)
Path difference $\displaystyle (\Delta)=\lambda$
$\displaystyle \phi=\frac{2 \pi \Delta}{\lambda}$
$\displaystyle \phi=2 \pi$
$\displaystyle I=4 I_{0} \cos ^{2} \frac{\phi}{2}$
$\displaystyle K=4 I_{0} \cos ^{2} \pi=4 I_{0}$
Path difference $\displaystyle =\frac{\lambda}{6}$
$\displaystyle \phi=\pi / 3$
$\displaystyle I=4 I_{0} \cos ^{2} \frac{\pi}{6}$
$\displaystyle =4 I_{0} \times \frac{3}{4}$
$\displaystyle =\frac{3}{4} K$
(i)
The magnification obtained by eye-piece lens $\displaystyle m_{e}=\left(1+\frac{D}{f_{e}}\right)$
The magnification obtained by objective lens $\displaystyle m_{0}=\frac{v_{0}}{-u_{0}}$
Hence the total magnifying power is:
$\displaystyle m=m_{0} \times m_{e}$
$\displaystyle =\frac{v_{0}}{-u_{0}}\left(1+\frac{D}{f_{e}}\right)$
(ii)
$\displaystyle m=\frac{f_{0}}{f_{e}}$
$\displaystyle f_{0}=100 \mathrm{~cm}$
$\displaystyle f_{e}=5 \mathrm{~cm}$
$\displaystyle m=\frac{100}{5}=20$
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