Question

In the figure, a ray of light is incident on a transparent liquid contained in a thin glass box at an angle of 45° with its one face. The emergent ray passes along the face AB. Find the refractive index of the liquid.

Hint:

The critical angle condition and refractive index relationship are key to understanding the behavior of light in different mediums, especially in designing optical instruments and understanding phenomena like total internal reflection.

Answer 1

Given:  Angle of incidence on the glass box: \(45^\circ\)  The emergent ray passes along the face AB, indicating that the angle of refraction at the glass-liquid interface is \(90^\circ\).  1. First Surface (Air-Glass Interface): \[ \frac{\sin 45^\circ}{\sin \theta} = \mu \] Given \(\sin 45^\circ = \frac{1}{\sqrt{2}}\): \[ \frac{1}{\sqrt{2}} = \mu \sin \theta \] 2. Second Surface (Glass-Liquid Interface): \[ \frac{\sin(90^\circ - \theta)}{\sin 90^\circ} = \frac{1}{\mu} \] Since \(\sin(90^\circ - \theta) = \cos \theta\) and \(\sin 90^\circ = 1\): \[ \cos \theta = \frac{1}{\mu} \] 3. Combining the Equations: \[ \frac{1}{\sqrt{2} \sin \theta} = \frac{1}{\cos \theta} \] Simplifying: \[ \cos \theta = \sqrt{2} \sin \theta \] \[ \tan \theta = \frac{1}{\sqrt{2}} \] 4. Finding \(\sin \theta\): From the triangle GEF: \[ \sin \theta = \frac{1}{\sqrt{3}} \] 5. Calculating the Refractive Index (\(\mu\)): \[ \mu = \frac{1}{\sqrt{2} \sin \theta} = \frac{1}{\sqrt{2} \times \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{3}{2}} \] Therefore, the refractive index of the liquid is: \[ \boxed{\sqrt{\frac{3}{2}}} \]