Question

The critical angle for a denser refractive index is 45°. The speed of light in water medium is \( 3 \times 10^8 \) m/s. The speed of light in the denser medium is:

• A. \( 2.12 \times 10^7 \, \text{m/s} \)
• B. \( 5 \times 10^7 \, \text{m/s} \)
• C. \( 3.12 \times 10^7 \, \text{m/s} \)
• D. \( \sqrt{5} \times 10^7 \, \text{m/s} \)

Hint:

The critical angle can help you determine the speed of light in the denser medium using the formula \( v = \frac{c}{\sin \theta_c} \).

Answer 1

The Correct Option is A

We know the critical angle \( \theta_c \) is related to the refractive indices of the two media: \[ \sin \theta_c = \frac{n_2}{n_1} \] Where \( n_1 \) is the refractive index of the denser medium (water) and \( n_2 \) is that of the rarer medium (air). Given \( \theta_c = 45^\circ \) and using \( n_1 = \frac{c}{v} \), where \( c \) is the speed of light in vacuum and \( v \) is the speed of light in the denser medium, we calculate: \[ v = \frac{c}{\sin \theta_c} = \frac{3 \times 10^8}{\sin 45^\circ} = 2.12 \times 10^7 \, \text{m/s} \] Thus, the speed of light in the denser medium is \( 2.12 \times 10^7 \, \text{m/s} \).