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Question 15 Marks
In an experiment with Foucault's apparatus, the various distances used are as follows:
Distance between the rotating and the fixed mirror $= 16m$
Distance between the lens and the rotating mirror $= 6m$
Distance between the source and the lens $= 2m$
When the mirror is rotated at a speed of 356 revolutions per second, the image shifts by 0.7mm. Calculate the speed of light from these data.
Answer
In the given Focault experiment, R = Distance between fixed and rotating mirror = 16m$\omega$ = Angular speed $= 356\text{rev/ s} = 356 × 2\pi \text{ rad/ sec}$
$b =$ Distance between lens and rotating mirror = 6m
$a =$ Distance between source and lens = 2m
$s =$ shift in image $= 0.7cm = 0.7 \times 10^{-3}m$
So, speed of light is given by,
$\text{C}=\frac{4\text{R}^2\omega\text{a}}{\text{s}(\text{R}+\text{b})}=\frac{4\times16^2\times356\times2\pi\times2}{0.7\times10^{-3}(16+6)}=2.975\times10^8\text{m/s}$
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Question 25 Marks
In an experiment to measure the speed of light by Fizeau's apparatus, following data are used:
Distance between the mirrors = 12.0km,
Number of teeth in the wheel = 180.
Find the minimum angular speed of the wheel for which the image is not seen.
Answer
In the given "Fizeau’' apparatus, $D = 12km = 12 \times 10^3m$
$n = 180 c = 3 \times 10^8m/sec$
We know, $\text{c}=\frac{2\text{Dn}\omega}{\pi}$
$\Rightarrow\omega=\frac{\pi\text{c}}{2\text{Dn}}\text{rad/sec}=\frac{\pi\text{c}}{2\text{Dn}}\times\frac{180}{\pi}\text{deg/sec}$
$\Rightarrow\omega=\frac{180\times3\times10^8}{24\times10^3\times180}=1.25\times10^4\text{deg/sec}$
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