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: Use travel times directly
Speed is inversely proportional to the time for a given distance. Light takes $t_0$ for $x$ in air, so for $4x$ in air it would take $4t_0$.

Step 2: Compare
In the medium the same $4x$ takes $t_1$, so $\mu = \dfrac{t_1}{4t_0}$.

Step 3: Invert
$\sin C = \dfrac{1}{\mu} = \dfrac{4t_0}{t_1}$.

Step 4: Answer
$C = \sin^{-1}\dfrac{4t_0}{t_1}$, option (B).

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