A medium of refractive index `mu` is separated from air using a plane surface. When light ray is incident on this medium at an angle `70^(@)` then it is found that refracted and reflected ray suffer equal deviations in opposite directions. What is refractive index of material ?

Options

A. `2 sin 70^(@)`
B. `sin 70^(@)`
C. `2 sec 70^(@)`
D. `2 cot 70^(@)`

Explanation

To solve the problem, we need to find the refractive index \( \mu \) of the medium given that a light ray is incident at an angle of \( 70^\circ \) and that the refracted and reflected rays suffer equal deviations in opposite directions.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a light ray incident on a plane surface separating air and a medium with refractive index \( \mu \).
- The angle of incidence \( i = 70^\circ \).
- The angle of reflection \( r \) is equal to the angle of incidence due to the law of reflection, so \( r = 70^\circ \).

2. **Calculating the Deviations**:
- The deviation for the reflected ray \( \Delta_1 \) can be calculated as:
\[ \Delta_1 = 180^\circ - 2i = 180^\circ - 2 \times 70^\circ = 180^\circ - 140^\circ = 40^\circ \]
- The deviation for the refracted ray \( \Delta_2 \) is given by:
\[ \Delta_2 = i - r = 70^\circ - r \]
- Since both deviations are equal in magnitude but opposite in direction, we have:
\[ \Delta_1 = \Delta_2 \]
- Thus, we can set up the equation:
\[ 40^\circ = 70^\circ - r \]

3. **Finding the Angle of Refraction**:
- Rearranging the equation gives:
\[ r = 70^\circ - 40^\circ = 30^\circ \]

4. **Applying Snell's Law**:
- Snell's law states:
\[ n_1 \sin(i) = n_2 \sin(r) \]
- Here, \( n_1 = 1 \) (refractive index of air), \( i = 70^\circ \), \( n_2 = \mu \), and \( r = 30^\circ \).
- Plugging in the values:
\[ 1 \cdot \sin(70^\circ) = \mu \cdot \sin(30^\circ) \]

5. **Calculating the Refractive Index**:
- We know \( \sin(30^\circ) = \frac{1}{2} \), so we can rewrite the equation:
\[ \sin(70^\circ) = \mu \cdot \frac{1}{2} \]
- Rearranging gives:
\[ \mu = 2 \cdot \sin(70^\circ) \]

6. **Final Calculation**:
- Using a calculator or trigonometric tables, we find \( \sin(70^\circ) \approx 0.9397 \).
- Therefore:
\[ \mu \approx 2 \cdot 0.9397 \approx 1.8794 \]

### Conclusion:
The refractive index \( \mu \) of the material is approximately \( 1.88 \).