Light ray is travelling from a denser medium into a rarer medium. The velocity of light in the denser and rarer medium is `2 xx 10^(8)` m/sec and `2.5 xx 10^(8)` m / sec. The critical angle of the two media is

Options

A. `sin^(-1)(5/4)`
B. `sin^(-1)(4/5)`
C. `sin^(-1)(1/2)`
D. `sin^(-1)(3/5)`

Explanation

To find the critical angle for light traveling from a denser medium to a rarer medium, we can follow these steps:

### Step 1: Identify the velocities of light in both media
- Given:
- Velocity of light in the denser medium, \( v_d = 2 \times 10^8 \) m/s
- Velocity of light in the rarer medium, \( v_r = 2.5 \times 10^8 \) m/s

### Step 2: Calculate the refractive indices of both media
- The refractive index \( \mu \) is given by the formula:
\[ \mu = \frac{c}{v} \]
where \( c \) is the speed of light in vacuum (approximately \( 3 \times 10^8 \) m/s).

- For the denser medium:
\[ \mu_d = \frac{c}{v_d} = \frac{3 \times 10^8}{2 \times 10^8} = \frac{3}{2} \]

- For the rarer medium:
\[ \mu_r = \frac{c}{v_r} = \frac{3 \times 10^8}{2.5 \times 10^8} = \frac{3}{2.5} = \frac{6}{5} \]

### Step 3: Apply Snell's Law to find the critical angle
- Snell's Law states:
\[ \mu_d \sin C = \mu_r \sin 90^\circ \]
Since \( \sin 90^\circ = 1 \), we can simplify this to:
\[ \mu_d \sin C = \mu_r \]

- Rearranging gives:
\[ \sin C = \frac{\mu_r}{\mu_d} \]

### Step 4: Substitute the values of the refractive indices
- Substituting the values we calculated:
\[ \sin C = \frac{\frac{6}{5}}{\frac{3}{2}} = \frac{6}{5} \times \frac{2}{3} = \frac{12}{15} = \frac{4}{5} \]

### Step 5: Calculate the critical angle
- To find the critical angle \( C \), we take the inverse sine:
\[ C = \sin^{-1}\left(\frac{4}{5}\right) \]

### Final Result
- The critical angle \( C \) is:
\[ C \approx 53.13^\circ \]