Total Internal Reflection: Practice Problem & Solution
The speed of light in media $M_1$ and $M_2$ is $1.5 \times 10^8 m/s$ and $2.0 \times 10^8 m/s$ respectively. A ray of light enters from medium $M_1$ to $M_2$ at an incidence angle $i$. If the ray suffers total internal reflection, the value of $i$ is: (2010 Mains)
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:
Critical angle is $\sin \theta_c = \frac{\mu_2}{\mu_1} = \frac{v_1}{v_2} = \frac{1.5 \times 10^8}{2.0 \times 10^8} = \frac{3}{4}$. For total internal reflection, the angle of incidence must be greater than or equal to the critical angle, so $i \ge \sin^{-1}\left(\frac{3}{4}\right)$.
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