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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° - Physics


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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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.
\[x = \frac{1  m}{\cos  30^\circ} = \frac{2}{\sqrt{3}}  m\]
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

Concept: Ray Optics - Mirror Formula
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HC Verma Class 11, 12 Concepts of Physics 1
Chapter 18 Geometrical Optics
Q 14 | Page 413
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