Question

Light travels a distance 'x' in time '\(t_0\)' in air and '\(4x\)' in time '\(t_1\)' in another denser medium. The critical angle for this medium is

• A. \(sin^{-1}(\frac{t_1}{t_0})\)
• B. \(sin^{-1}(\frac{4t_0}{t_1})\)
• C. \(sin^{-1}(\frac{4t_1}{t_0})\)
• D. \(sin^{-1}(\frac{t_0}{4t_1})\)

Hint:

Refractive index is the ratio of speeds, and the critical angle satisfies sin C = 1/mu.

Answer 1

The Correct Option is B

Step 1: Understand the concept
The refractive index of the medium is \(\mu = \dfrac{c}{v}\), where \(c\) is the speed in air and \(v\) is the speed in the medium. The critical angle satisfies \(\sin C = \dfrac{1}{\mu}\).

Step 2: Find the speeds
In air: \(c = \dfrac{x}{t_0}\). In the medium: \(v = \dfrac{4x}{t_1}\).

Step 3: Find \(\mu\)
\[ \mu = \frac{c}{v} = \frac{x/t_0}{4x/t_1} = \frac{t_1}{4t_0} \]

Step 4: Critical angle
\[ \sin C = \frac{1}{\mu} = \frac{4t_0}{t_1} \Rightarrow C = \sin^{-1}\left(\frac{4t_0}{t_1}\right) \]
Option (B). The medium is denser, so \(t_1 > 4t_0\) and the ratio is less than 1, so the inverse sine is defined.

Final Answer:
The critical angle is arcsin(4 t0 / t1). This is option (B). \[ \boxed{\text{(B) }\sin^{-1}\left(\frac{4t_0}{t_1}\right)} \]