Question

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

• A. $sin \theta =\frac{1}{n^{2}}$
• B. $sin \theta =\frac{1}{n_{1}}$
• C. $sin \theta =\frac{n_{2}}{n_{1}}$
• D. $sin \theta =\frac{n_{1}}{n_{2}}$

Answer 1

The Correct Option is B

From snell?s law
$n_{1} sin \theta =1$
$ sin \theta =\frac{1}{n_{1}}$

Answer 2

When a ray of light passes from one medium to another, it undergoes refraction according to Snell's law, which states:
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
where:
- \( n_1 \) and \( n_2 \) are the refractive indices of the two media,
- \( \theta_1 \) is the angle of incidence,
- \( \theta_2 \) is the angle of refraction.
Given that the ray is incident at an angle \( \theta \) on the surface of separation of slabs A and B, and after refraction from B into air grazes the surface of B, it means the angle of refraction is 90 degrees. This is because the ray grazes the surface of B, which implies it makes an angle of 90 degrees with the surface.
So, applying Snell's law, we have:
\[ n_1 \sin(\theta) = n_2 \sin(90^\circ) \]
Since \( \sin(90^\circ) = 1 \), the equation simplifies to:
\[ n_1 \sin(\theta) = n_2 \]
Dividing both sides by \( n_1 \), we get:
\[ \sin(\theta) = \frac{n_2}{n_1} \]
Given that the refractive index of air (\( n_2 \)) is approximately 1, the equation becomes:
\[ \sin(\theta) = \frac{1}{n_1} \]
Therefore,  The correct Answer is Option (B) \( \sin(\theta) = \frac{1}{n_1} \).