A and B are two parallel sided transparent slabs of refractive indices `n_(1) " and" n_(2)` respectively. A ray is incident at an angle `theta` on the surface of separation of A and B and after refraction from B into air grazes the surface of B. Then

Options

A. `sin theta = (1)/(n_(2))`
B. `sin theta = (1)/(n_(1))`
C. `sin theta = (n_(2))/(n_(1))`
D. `sin theta = (n_(1))/(n_(2))`

Explanation

To solve the problem, we need to analyze the refraction of light as it passes through two transparent slabs (A and B) with different refractive indices and then exits into air. We will apply Snell's law to find the relationship between the angles of incidence and refraction.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Refractive index of slab A: \( n_1 \)
- Refractive index of slab B: \( n_2 \)
- Angle of incidence on the surface of separation: \( \theta \)

2. **Apply Snell's Law at the Interface of Slab A and Slab B:**
- When the ray of light strikes the surface separating slabs A and B, we can use Snell's law:
\[ n_1 \sin(\theta) = n_2 \sin(\theta_2) \]
- Here, \( \theta_2 \) is the angle of refraction in slab B.

3. **Determine the Condition for Grazing Refraction:**
- The problem states that the ray grazes the surface of slab B when it exits into air. This means that the angle of refraction at the interface between slab B and air is \( 90^\circ \).
- According to Snell's law at the interface of slab B and air:
\[ n_2 \sin(\theta_2) = n_{air} \sin(90^\circ) \]
- Since \( \sin(90^\circ) = 1 \) and the refractive index of air \( n_{air} \) is approximately 1, we can write:
\[ n_2 \sin(\theta_2) = 1 \]
- Therefore, we can express \( \sin(\theta_2) \) as:
\[ \sin(\theta_2) = \frac{1}{n_2} \]

4. **Substitute \( \sin(\theta_2) \) Back into Snell's Law:**
- From the first application of Snell's law, we have:
\[ n_1 \sin(\theta) = n_2 \sin(\theta_2) \]
- Substituting \( \sin(\theta_2) \):
\[ n_1 \sin(\theta) = n_2 \left(\frac{1}{n_2}\right) \]
- This simplifies to:
\[ n_1 \sin(\theta) = 1 \]

5. **Final Relation:**
- Rearranging gives us:
\[ \sin(\theta) = \frac{1}{n_1} \]
- This equation describes the relationship between the angle of incidence \( \theta \) and the refractive index \( n_1 \) of slab A.

### Summary:
The final relation derived from the conditions given in the problem is:
\[ \sin(\theta) = \frac{1}{n_1} \]