Two transparent media A and B are separated by a plane boundary. The speed of light in medium A is `2.0xx10^(8)ms^(-1)` and in medium B is `2.5xx10^(8)ms^(-1)`. The critical angle for which a ray of light going from A to B suffers total internal reflection is

Options

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

Answer

To find the critical angle for total internal reflection when light travels from medium A to medium B, we can follow these steps:

### Step 1: Identify the given values
- Speed of light in medium A, \( V_A = 2.0 \times 10^8 \, \text{m/s} \)
- Speed of light in medium B, \( V_B = 2.5 \times 10^8 \, \text{m/s} \)

### Step 2: Calculate the refractive indices
The refractive index \( n \) of a medium is given by the formula:
\[ n = \frac{c}{V} \]
where \( c \) is the speed of light in vacuum (approximately \( 3.0 \times 10^8 \, \text{m/s} \)).

#### For medium A:
\[ n_A = \frac{c}{V_A} = \frac{3.0 \times 10^8}{2.0 \times 10^8} = \frac{3}{2} = 1.5 \]

#### For medium B:
\[ n_B = \frac{c}{V_B} = \frac{3.0 \times 10^8}{2.5 \times 10^8} = \frac{3}{2.5} = \frac{30}{25} = \frac{6}{5} = 1.2 \]

### Step 3: Use Snell's Law to find the critical angle
The critical angle \( \theta_c \) can be found using Snell's Law:
\[ n_A \sin(\theta_c) = n_B \sin(90^\circ) \]
Since \( \sin(90^\circ) = 1 \), we have:
\[ n_A \sin(\theta_c) = n_B \]

Substituting the values:
\[ 1.5 \sin(\theta_c) = 1.2 \]

### Step 4: Solve for \( \sin(\theta_c) \)
Rearranging the equation gives:
\[ \sin(\theta_c) = \frac{1.2}{1.5} = \frac{12}{15} = \frac{4}{5} \]

### Step 5: Calculate the critical angle
To find the critical angle \( \theta_c \):
\[ \theta_c = \sin^{-1}\left(\frac{4}{5}\right) \]

### Final Answer
The critical angle for which a ray of light going from medium A to medium B suffers total internal reflection is:
\[ \theta_c = \sin^{-1}\left(\frac{4}{5}\right) \]