Question

The refractive index of the material of a prism is $1.5$ and the angle of the prism is $6°$. find the deviation produced by the prism for a small angle of incidence.

Answer 1

Step 1: Given parameters

Refractive index of the prism, $n=1.5$

Angle of the prism, $A=6°$

Step 2: Formula used

The angle of deviation can be found by using the prism formula:

$n=\frac{\sin \left({\displaystyle \frac{A+{\delta }_m}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}$

Where, ${\delta }_m$ is the angle of deviation, $A$ is the angle of the prism, and $n$ is the refractive index.

We know that for small angles, $\sin \theta =\theta$.

This implies that,

$n=\frac{A+{\delta }_m}{A}...\left(1\right)$

Step 3: Calculate the angle of deviation

Substitute $n=1.5$, and, $A=6°$ in equation (1).

$1.5=\frac{6+{\delta }_m}{6}$

${\delta }_m=3°$

Hence, the angle of deviation produced by the prism is $3°$.