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प्रश्न
A light ray falling at an angle of 45° with the surface of a clean slab of ice of thickness 1.00 m is refracted into it at an angle of 30°. Calculate the time taken by the light rays to cross the slab. Speed of light in vacuum = 3 × 108 m s−1.
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उत्तर
Given,
Angle of incidence, i = 45°
Angle of refraction, r = 30°
Using Snell's law,
\[\frac{\sin i}{\sin r} = \frac{3 \times {10}^8}{v}\]
\[= \frac{\sin 45^\circ}{\sin 30^\circ } = \frac{\left( \frac{1}{\sqrt{2}} \right)}{\left( \frac{1}{2} \right)}\]
\[= \frac{2}{\sqrt{2}} = \sqrt{2}\]
\[So, \]
\[ v = \frac{3 \times {10}^8}{\sqrt{2}} m/s\]
Let x be the distance travelled by light in the slab.
Now,
\[x = \frac{1 m}{\cos 30^\circ} = \frac{2}{\sqrt{3}} m\]
We know:
Time taken
\[= \frac{Distance}{Speed}\]
\[= \frac{2}{\sqrt{3}} \times \frac{\sqrt{2}}{3 \times {10}^8}\]
= 0.54 × 10−8
= 5.4 × 10−9 s
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