The velocities of light in two different mediums are `2xx10^(8) ms^( -1)` and `2.5xx10^(8) ms^(-1)` respectively. The critical angle for there mediums is

Options

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

Explanation

To find the critical angle for the two mediums given their velocities of light, we can follow these steps:

### Step 1: Identify the velocities of light in the two mediums
Let the velocity of light in medium 1 (M1) be \( V_1 = 2 \times 10^8 \, \text{m/s} \) and in medium 2 (M2) be \( V_2 = 2.5 \times 10^8 \, \text{m/s} \).

### Step 2: Calculate the refractive indices of the two mediums
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 \( 3 \times 10^8 \, \text{m/s} \)).

For medium 1:
\[ n_1 = \frac{c}{V_1} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5 \]

For medium 2:
\[ n_2 = \frac{c}{V_2} = \frac{3 \times 10^8}{2.5 \times 10^8} = 1.2 \]

### Step 3: Apply the formula for critical angle
The critical angle \( \theta_c \) is given by Snell's law at the point of total internal reflection:
\[ n_1 \sin \theta_c = n_2 \sin 90^\circ \]
Since \( \sin 90^\circ = 1 \), we can simplify this to:
\[ n_1 \sin \theta_c = n_2 \]

### Step 4: Solve for \( \sin \theta_c \)
Rearranging the equation gives:
\[ \sin \theta_c = \frac{n_2}{n_1} \]
Substituting the values of \( n_1 \) and \( n_2 \):
\[ \sin \theta_c = \frac{1.2}{1.5} = \frac{4}{5} \]

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

### Final Answer
Thus, the critical angle for the two mediums is:
\[ \theta_c = \sin^{-1}\left(\frac{4}{5}\right) \]

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