Total Internal Reflection: Practice Problem & Solution
A disc is placed on a surface of pond which has refractive index $\frac{5}{3}$. A source of light is placed $4 m$ below the surface of liquid. The minimum radius of disc will be so light is not coming out: (2001)
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:
The light is cut off if the disc covers the area defined by the critical angle. Radius is $r = \frac{h}{\sqrt{\mu^2 - 1}}$. Substituting $h = 4 m$ and $\mu = \frac{5}{3}$, $r = \frac{4}{\sqrt{(\frac{5}{3})^2 - 1}} = \frac{4}{\frac{4}{3}} = 3 m$.
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