Question

The refractive index of a transparent liquid filled in an equilateral hollow prism is √2. The angle of minimum deviation for the liquid will be____.

Answer 1

Correct Answer: 30

We are given the refractive index of a liquid and the angle of an equilateral prism. We need to find the angle of minimum deviation.

Solution

1. Formula for Refractive Index at Minimum Deviation:

The refractive index \( \mu \) of the prism material is given by:

\( \mu = \frac{\sin \left( \frac{A + \delta_{\text{min}}}{2} \right)}{\sin \left( \frac{A}{2} \right)} \)

where:

• \( \mu \) is the refractive index
• \( A \) is the angle of the prism
• \( \delta_{\text{min}} \) is the angle of minimum deviation

2. Substitute Given Values:

Given \( \mu = \sqrt{2} \) and \( A = 60^\circ \), we have:

\( \sqrt{2} = \frac{\sin \left( \frac{60^\circ + \delta_{\text{min}}}{2} \right)}{\sin \left( \frac{60^\circ}{2} \right)} \)

\( \sqrt{2} = \frac{\sin \left( 30^\circ + \frac{\delta_{\text{min}}}{2} \right)}{\sin(30^\circ)} \)

3. Solve for \( \delta_{\text{min}} \):

We know \( \sin(30^\circ) = \frac{1}{2} \), so:

\( \sqrt{2} = \frac{\sin \left( 30^\circ + \frac{\delta_{\text{min}}}{2} \right)}{\frac{1}{2}} \)

\( \sin \left( 30^\circ + \frac{\delta_{\text{min}}}{2} \right) = \frac{\sqrt{2}}{2} \)

\( 30^\circ + \frac{\delta_{\text{min}}}{2} = \sin^{-1} \left( \frac{\sqrt{2}}{2} \right) = 45^\circ \)

\( \frac{\delta_{\text{min}}}{2} = 45^\circ - 30^\circ = 15^\circ \)

\( \delta_{\text{min}} = 2 \times 15^\circ = 30^\circ \)

Final Answer

Thus, the angle of minimum deviation \( \delta_{\text{min}} \) is \( 30^\circ \).