Velocity of light in a liquid is `1.5 xx 10^(8) m//s` and in air, it is `3 xx 10^(8)m//s`. If a ray of light passes from this liquid to air, calculate the value of critical angle.

Explanation

To find the critical angle when light passes from a liquid to air, we can follow these steps:

### Step 1: Understand the relationship between velocity of light and refractive index
The refractive index (μ) of a medium is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in the medium. The formula is given by:

\[ \mu = \frac{c}{v} \]

where:
- \( c \) is the speed of light in air (or vacuum),
- \( v \) is the speed of light in the medium (liquid in this case).

### Step 2: Calculate the refractive index of the liquid
Given:
- Velocity of light in air, \( c = 3 \times 10^8 \, \text{m/s} \)
- Velocity of light in liquid, \( v = 1.5 \times 10^8 \, \text{m/s} \)

Using the formula for refractive index:

\[ \mu = \frac{c}{v} = \frac{3 \times 10^8}{1.5 \times 10^8} \]

Calculating this gives:

\[ \mu = \frac{3}{1.5} = 2 \]

### Step 3: Use the formula for critical angle
The critical angle (C) can be found using Snell's law, which states:

\[ \mu_1 \sin C = \mu_2 \sin 90^\circ \]

Here, \( \mu_1 \) is the refractive index of the liquid (2), and \( \mu_2 \) is the refractive index of air (approximately 1). Since \( \sin 90^\circ = 1 \), the equation simplifies to:

\[ 2 \sin C = 1 \]

### Step 4: Solve for the critical angle
Rearranging the equation gives:

\[ \sin C = \frac{1}{2} \]

To find the angle \( C \), we take the inverse sine:

\[ C = \sin^{-1}\left(\frac{1}{2}\right) \]

This results in:

\[ C = 30^\circ \]

### Conclusion
Thus, the critical angle when light passes from the liquid to air is:

\[ \text{Critical angle } C = 30^\circ \]

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