Refractive index of water with respect to air is `sqrt(2)`. Alight ray is incident on the surface at an angle 30° travelling through water. The deviation of light ray is

Options

A. `30^@`
B. `120^@`
C. `15^@`
D. `60^@`

Explanation

To solve the problem, we need to find the deviation of a light ray when it travels from water to air, given the refractive index of water with respect to air is \(\sqrt{2}\) and the angle of incidence is \(30^\circ\).

### Step-by-Step Solution:

1. **Identify the Refractive Indices**:
- The refractive index of water with respect to air (\(\mu_{water/air}\)) is given as \(\sqrt{2}\).
- The refractive index of air (\(\mu_{air}\)) is \(1\).
- Therefore, the refractive index of water (\(\mu_{water}\)) can be expressed as:
\[ \mu_{water} = \sqrt{2} \times \mu_{air} = \sqrt{2} \times 1 = \sqrt{2} \]

2. **Apply Snell's Law**:
- Snell's Law states that:
\[ \mu_1 \sin i = \mu_2 \sin r \]
- Here, \(\mu_1 = \mu_{water} = \sqrt{2}\), \(\mu_2 = \mu_{air} = 1\), and the angle of incidence \(i = 30^\circ\).
- Substitute the values into Snell's Law:
\[ \sqrt{2} \sin(30^\circ) = 1 \sin(r) \]

3. **Calculate \(\sin(30^\circ)\)**:
- We know that \(\sin(30^\circ) = \frac{1}{2}\).
- Substitute this value into the equation:
\[ \sqrt{2} \cdot \frac{1}{2} = \sin(r) \]
- This simplifies to:
\[ \sin(r) = \frac{\sqrt{2}}{2} \]

4. **Find the Angle of Refraction \(r\)**:
- The value \(\sin(r) = \frac{\sqrt{2}}{2}\) corresponds to:
\[ r = 45^\circ \]

5. **Calculate the Deviation**:
- The deviation (\(\Delta\)) is given by the formula:
\[ \Delta = r - i \]
- Substitute the values of \(r\) and \(i\):
\[ \Delta = 45^\circ - 30^\circ = 15^\circ \]

6. **Final Answer**:
- The deviation of the light ray is \(15^\circ\).