Question

Light travels in two media \(M_1\) and \(M_2\) with speeds \(1.5 × 10^8ms^{–1 }\)and \(2.0 × 10^8ms^{–1}\), respectively. The critical angle between them is:

• A. \(tan^{-1}\bigg(\frac{3}{\sqrt7}\bigg)\)
• B. \(tan^{-1}\bigg(\frac{2}{3}\bigg)\)
• C. \(cos^{-1}\bigg(\frac{3}{4}\bigg)\)
• D. \(sin^{-1}\bigg(\frac{2}{3}\bigg)\)

Answer 1

The Correct Option is A

To find the critical angle between two media \(M_1\) and \(M_2\) where the speeds of light are given as \(1.5 \times 10^8 \, \text{m/s}\) and \(2.0 \times 10^8 \, \text{m/s}\) respectively, we start by determining the refractive indices of each medium. The refractive index \(n\) of a medium is inversely proportional to the speed of light in the medium and is given by the formula:

n = \frac{c}{v},

where \(c\) is the speed of light in a vacuum, approximately \(3.0 \times 10^8 \, \text{m/s}\).

Let's calculate the refractive indices \(n_1\) for \(M_1\) and \(n_2\) for \(M_2\):

1. For medium \(M_1\): n_1 = \frac{3.0 \times 10^8}{1.5 \times 10^8} = 2.
2. For medium \(M_2\): n_2 = \frac{3.0 \times 10^8}{2.0 \times 10^8} = 1.5.

To find the critical angle \(\theta_c\), we use the formula for the critical angle when light travels from a medium with a higher refractive index (\(n_1\)) to a medium with a lower refractive index (\(n_2\)):

\sin(\theta_c) = \frac{n_2}{n_1}.

Substituting the values, we get:

\sin(\theta_c) = \frac{1.5}{2} = 0.75.

The critical angle \(\theta_c\) can be found using the inverse sine function:

\theta_c = \sin^{-1}(0.75).

This matches with the given options described by their trigonometric inverse functions. The correct answer given was:

\theta_c = \tan^{-1}\left(\frac{3}{\sqrt{7}}\right).

However, since \(\theta_c = \sin^{-1}(0.75)\), which is approximately equal, the explanation is coherent with the given multiple-choice option \(\tan^{-1}\left(\frac{3}{\sqrt{7}}\right)\).