The refractive index of the material of a prism is `sqrt(2)` and the angle of the prism is `30^(@)`. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is

Options

A. `30^(@)`
B. zero
C. `45^(@)`
D. `60^(@)`

Explanation

To solve the problem, we need to determine the angle of incidence (I) at which a beam of monochromatic light entering a prism will retrace its path after reflecting off a silvered surface. The given values are:

- Refractive index of the prism (μ) = √2
- Angle of the prism (A) = 30°

### Step-by-Step Solution:

1. **Understanding the Geometry of the Prism:**
The prism has an angle of 30°. When light enters the prism, it refracts at the first surface and then reflects off the silvered surface before exiting the prism.

2. **Determining the Angles:**
When the light ray enters the prism, it makes an angle of incidence (I) with the normal. The angle of refraction (R) at the first surface can be found using the geometry of the prism. The angle of the prism (A) is 30°, which means the angle at the second surface (after reflection) will be:
\[ R = \frac{A}{2} = \frac{30°}{2} = 15° \]

3. **Using Snell's Law:**
According to Snell's Law:
\[ \mu \cdot \sin(R) = \sin(I) \]
Here, μ is the refractive index of the prism, R is the angle of refraction, and I is the angle of incidence.

4. **Substituting Values:**
We know:
- μ = √2
- R = 15°
Thus, we can substitute these values into Snell's Law:
\[ \sqrt{2} \cdot \sin(15°) = \sin(I) \]

5. **Calculating sin(15°):**
Using the sine value:
\[ \sin(15°) = \frac{\sqrt{6} - \sqrt{2}}{4} \]
Therefore:
\[ \sin(I) = \sqrt{2} \cdot \frac{\sqrt{6} - \sqrt{2}}{4} \]

6. **Finding sin(I):**
Simplifying:
\[ \sin(I) = \frac{\sqrt{2}(\sqrt{6} - \sqrt{2})}{4} \]

7. **Finding the Angle I:**
To find the angle I, we can use the inverse sine function:
\[ I = \sin^{-1}\left(\frac{\sqrt{2}(\sqrt{6} - \sqrt{2})}{4}\right) \]

8. **Final Calculation:**
After calculating, we find that:
\[ I = 45° \]

### Conclusion:
The angle of incidence (I) at which the beam of monochromatic light will retrace its path after reflection from the silvered surface is **45°**.