A prism `(n = 2)` of apex angle `90^(@)` is palced in air `(n = 1)`. What should be the angle of incidence so that light ray strikes the second surface at an angle of incidence `60^(@)`.

Explanation

To solve the problem, we need to find the angle of incidence \( I \) such that a light ray strikes the second surface of a prism with an apex angle of \( 90^\circ \) at an angle of incidence of \( 60^\circ \). The refractive index of the prism is \( n = 2 \) and that of air is \( n = 1 \).

### Step-by-Step Solution:

1. **Understanding the Geometry of the Prism**:
- The prism has an apex angle \( A = 90^\circ \).
- The light ray enters the prism from air and strikes the first surface (AB) of the prism.

2. **Identifying Angles**:
- Let \( r \) be the angle of refraction at the first surface (AB).
- The angle of incidence at the second surface (BC) is given as \( 60^\circ \).

3. **Using the Prism Formula**:
- For a prism, the relationship between the angle of incidence \( I \), angle of refraction \( r \), and the angle of the prism \( A \) is given by:
\[ r + r' = A \]
- Here, \( r' \) is the angle of refraction at the second surface (BC).
- Since \( A = 90^\circ \), we have:
\[ r + r' = 90^\circ \]
- Given that \( r' = 60^\circ \), we can find \( r \):
\[ r + 60^\circ = 90^\circ \implies r = 30^\circ \]

4. **Applying Snell's Law at the First Surface (AB)**:
- Snell's Law states:
\[ n_1 \sin I = n_2 \sin r \]
- Here, \( n_1 = 1 \) (air), \( n_2 = 2 \) (prism), and \( r = 30^\circ \).
- Substituting the values:
\[ 1 \cdot \sin I = 2 \cdot \sin 30^\circ \]
- We know \( \sin 30^\circ = \frac{1}{2} \), so:
\[ \sin I = 2 \cdot \frac{1}{2} = 1 \]

5. **Finding the Angle of Incidence**:
- Since \( \sin I = 1 \), we find:
\[ I = 90^\circ \]

### Final Answer:
The angle of incidence \( I \) should be \( 90^\circ \).