Question

The angle of minimum deviation for a 75° prism of dense glass is found to be 45° when in air and 15° when immersed in certain liquid. The refractive index of the liquid is

Answer 1

Answer on Question #55301, Physics / Optics

The angle of minimum deviation for a $75{}^{\circ}$ prism of dense glass is found to be $45{}^{\circ}$ when in air and $15{}^{\circ}$ when immersed in certain liquid. The refractive index of the liquid is

Solution:

The minimum deviation $D$ in a prism occurs when the entering angle and the exiting angle are the same, a particularly symmetrical configuration. Applying Snell's Law at the interfaces you can derive the following relationship:

$$\frac {n}{n _ {0}} = \frac {\sin \frac {D _ {1} + A}{2}}{\sin \frac {A}{2}}$$

where $n$ is the refractive index of glass, $n_0 = 1$ is the refractive index of air, $D_1$ is the angle of minimum deviation, and $A$ is the internal angle of the prism.

Thus,

$$n = \frac {\sin \frac {4 5 {}^ {\circ} + 7 5 {}^ {\circ}}{2}}{\sin \frac {7 5 {}^ {\circ}}{2}} = 1. 4 2 3$$

Equation for liquid is

$$\frac {n}{n _ {L}} = \frac {\sin \frac {D _ {2} + A}{2}}{\sin \frac {A}{2}}$$

Thus,

$$n _ {L} = n \frac {\sin \frac {A}{2}}{\sin \frac {D _ {2} + A}{2}} = 1. 4 2 3 \cdot \frac {\sin \frac {7 5 {}^ {\circ}}{2}}{\sin \frac {7 5 {}^ {\circ} + 1 5 {}^ {\circ}}{2}} = 1. 2 2 5$$

Answer: $n_{L} = 1.225$

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