Thin symmetric prism of refractive index µ = 1.5. Find ratio of incident angle and minimum deviation.

The correct answer is 3/2.

Step-by-Step Explanation:

For a thin prism of small angle $A$ and refractive index $\mu$, the angle of minimum deviation ${\delta }_m$ is given by the formula:
${\delta }_m=(\mu -1)A$

In a symmetric prism at the minimum deviation position, the refraction is symmetric, meaning the angle of refraction at the first surface, $r_1$, is equal to the angle of refraction at the second surface, $r_2$:
$r_1=r_2=\frac{A}{2}$

By Snell's Law at the first refracting interface:
$sini=\mu sin{r}_1$

Since the prism is thin, all angles (including the angle of incidence $i$ and the angle of refraction $r_1$) are very small. Thus, we can use the small-angle approximation $sin\theta \approx \theta$:
$i\approx \mu r_1=\mu \left(\frac{A}{2}\right)$

Now, let's find the ratio of the angle of incidence $i$ to the angle of minimum deviation ${\delta }_m$:
$\frac{i}{{\delta }_m}=\frac{\mu \left(\frac{A}{2}\right)}{(\mu -1)A}=\frac{\mu }{2(\mu -1)}$

Given the refractive index of the prism is $\mu =1.5$, substitute this value into our ratio equation:
$\frac{i}{{\delta }_m}=\frac{1.5}{2(1.5-1)}$
$\frac{i}{{\delta }_m}=\frac{1.5}{2\left(0.5\right)}$
$\frac{i}{{\delta }_m}=\frac{1.5}{1}=\frac{3}{2}$

Thus, the ratio of the angle of incidence to the angle of minimum deviation is 3/2.