Question

A rectangular glass slab ABCD (refractive index 1.5) is surrounded by a transparent liquid (refractive index 1.25) as shown in the figure. A ray of light is incident on face AB at an angle \(i\) such that it is refracted out grazing the face AD. Find the value of angle \(i\).

Hint:

For grazing refraction, the angle of refraction is \(90^\circ\), and Snell's law can be used to determine the angle of incidence in the first medium.

Answer 1

To solve this problem, Snell's law is applied, which establishes a relationship between the angles of incidence and refraction at the boundary of two media: \[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \] Where:
- \(n_1\) and \(n_2\) represent the refractive indices of the two media.
- \(\theta_1\) is the angle of incidence in the first medium.
- \(\theta_2\) is the angle of refraction in the second medium.
Step 1: Conditions for Grazing Refraction At the interface between the glass and the liquid, the refracted ray grazes the face AD. This means the angle of refraction at this interface must be \(90^\circ\).
Step 2: Applying Snell’s Law at the Glass-Liquid Interface The following values are given:
- The refractive index of glass: \(n_{\text{glass}} = 1.5\)
- The refractive index of the liquid: \(n_{\text{liquid}} = 1.25\)
- The angle of refraction: \( \theta_2 = 90^\circ \) Snell's Law at the interface yields: \[ n_{\text{glass}} \sin i = n_{\text{liquid}} \sin 90^\circ \] \[ 1.5 \sin i = 1.25 \] \[ \sin i = \frac{1.25}{1.5} \] \[ \sin i = \frac{5}{6} \]
Step 3: Calculating the Angle The angle of incidence \(i\) is calculated as: \[ i = \sin^{-1} \left( \frac{5}{6} \right) \approx 56.44^\circ \] Therefore, the angle of incidence \(i\) is approximately \(56.44^\circ\).