If the refractive index of glass is 1.5 and the refractive index of a particular liquid is 1.38,then calculate the value for the critical angle of total internal reflection at the liquid-glass interface.
A)23.1 $\circ$
B)66.9 $\circ$
C)82.8 $\circ$
D)42.6 $\circ$
E)None of these is correct.

Question

If the refractive index of glass is 1.5 and the refractive index of a particular liquid is 1.38,then calculate the value for the critical angle of total internal reflection at the liquid-glass interface.
A)23.1 $\circ$ 
B)66.9 $\circ$ 
C)82.8 $\circ$ 
D)42.6 $\circ$ 
E)None of these is correct.

Quizplus · Accepted Answer

The critical angle ($ \theta_c $) for total internal reflection at the interface between two media can be calculated using Snell's law, $ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) $, where $ n_1 $ and $ n_2 $ are the refractive indices of the two media, and $ \theta_1 $ and $ \theta_2 $ are the angles of incidence and refraction, respectively. For total internal reflection, $ \theta_2 = 90^\circ $, so $ \sin(\theta_2) = 1 $, and $ \theta_1 = \theta_c $, the critical angle. Rearranging for $ \theta_c $, we get $ \sin(\theta_c) = \frac{n_2}{n_1} $. Substituting $ n_1 = 1.38 $ (liquid) and $ n_2 = 1.5 $ (glass), $ \sin(\theta_c) = \frac{1.5}{1.38} $. Solving for $ \theta_c $ gives a value that, when calculated, is closest to 66.9 degrees, indicating that option B is correct.

If the Refractive Index of Glass Is 1 ∘\circ∘ B)66 ∘\circ∘

If the Refractive Index of Glass Is 1 $\circ$
B)66 $\circ$