Total Internal Reflection: Practice Problem & Solution
A small source of light is $4 m$ below the surface of water of refractive index $5/3$. In order to cut off all the light, coming out of water surface, minimum diameter of the disc placed on the surface of water is (1994)
Solution Explained:
To solve this problem, we apply the core principles of Total Internal Reflection. Understanding the underlying formula is key to arriving at the correct answer below:
Radius of the disc is $r = \frac{h}{\sqrt{\mu^2 - 1}} = \frac{4}{\sqrt{(\frac{5}{3})^2 - 1}} = \frac{4}{4/3} = 3 m$. The minimum diameter of the disc required is $d = 2r = 2(3) = 6 m$.
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