One face of prism of refracting angle `30^@` and refractive index `1.414` is silvered. At what angle must a ray of light fall on the unsilvered face so that it retraces its path out of the prism ?

Explanation

To solve the problem, we need to determine the angle at which a ray of light must strike the unsilvered face of a prism with a refracting angle of \(30^\circ\) and a refractive index of \(1.414\) (which is \(\sqrt{2}\)), such that it retraces its path after reflecting off the silvered face.

### Step-by-Step Solution:

1. **Understanding the Geometry of the Prism**:
- The prism has a refracting angle \(A = 30^\circ\).
- One face of the prism is silvered, which means it acts as a mirror.

2. **Identifying the Angles**:
- Let \(R_1\) be the angle of incidence at the silvered face.
- Let \(R_2\) be the angle of refraction as the ray enters the prism.
- According to the prism's geometry, we have:
\[ R_1 + R_2 = A \]
- Since \(A = 30^\circ\), we can write:
\[ R_1 + R_2 = 30^\circ \]

3. **Using Snell's Law**:
- Snell's law states:
\[ n_1 \sin I = n_2 \sin R \]
- Here, \(n_1 = 1\) (the refractive index of air), \(n_2 = 1.414\) (the refractive index of the prism), \(I\) is the angle of incidence, and \(R\) is the angle of refraction.
- For the ray entering the prism:
\[ \sin I = 1.414 \sin R_2 \]

4. **Finding the Relationship Between Angles**:
- From the previous step, we know:
\[ R_1 = 30^\circ - R_2 \]
- Substituting this into Snell's law gives:
\[ \sin I = 1.414 \sin(30^\circ - R_1) \]

5. **Considering the Reflection**:
- For the ray to retrace its path, it must reflect off the silvered face at an angle equal to \(R_1\) (which is the angle of incidence).
- The angle of incidence at the silvered face must equal the angle of reflection, which is also \(R_1\).

6. **Finding the Angle of Incidence**:
- Since \(R_2 = 0^\circ\) when the ray retraces its path, we have:
\[ R_1 = 30^\circ \]
- Thus, we can apply Snell's law again:
\[ \sin I = 1.414 \sin(30^\circ) \]
- We know \(\sin(30^\circ) = \frac{1}{2}\):
\[ \sin I = 1.414 \times \frac{1}{2} = \frac{1.414}{2} \]

7. **Calculating the Angle of Incidence**:
- Therefore:
\[ \sin I = \frac{1.414}{2} = 0.707 \]
- Taking the inverse sine:
\[ I = \sin^{-1}(0.707) = 45^\circ \]

### Final Answer:
The angle at which the ray of light must fall on the unsilvered face of the prism is \(45^\circ\).