Question

(a) A ray 'PQ' of light is incident on the face AB of a glass prism ABC (as shown in the figure) and emerges out of the face AC. Trace the path of the ray. Show that $\angle i+\angle e=\angle A+\angle \delta$. When $\delta$ and e denote the angle of deviation and angle of emergence respectively.



Plot a graph showing the variation of the angle of deviation as a function of angle of incidence. State the condition under which $\delta$ is minimum.

(b) Find out the relation between the refractive index ($\mu$) of the glass prism and $\angle A$ for the case when the angle of prism (A) is equal to the angle of minimum deviation $\left({\delta }_m\right)$. Hence obtain the value of the refractive index for angle of prism $A={60}^{\circ }$.

Answer 1

(a) Let the incident ray meet refracting face AB of the prism at point P. Ray PQ is the refracted ray inside the prism and ${\delta }_2$ and $r_1$ are the angle of the deviation and refraction at interface AB. At interface AC the ray goes out of the prism. Let e be the angle of emergence. The angle of deviation at point Q is ${\delta }_2$ as shown in figure.



Using geometry, we see that at point P,

$i={\delta }_1+r_1$ $\therefore {\delta }_1=i-r_1$

and at point Q, $e={\delta }_1+r_2$

${\delta }_2=e-r_2$

The total deviation $\delta$, suffered by the incident ray is equal to ${\delta }_1+{\delta }_2$.

or $\delta ={\delta }_1+{\delta }_2$
$=(i-r_1)+(e-r_2)$
$=(i+e)-(r_1+r_2)$ ...(i)

In quadrilateral POQA, the sum of all four angle is ${360}^{\circ }$.

$P+O+Q+A={360}^{\circ }$

as P and Q both are right angles

$P+Q={180}^{\circ }$

$O+A={180}^{\circ }$ ...(ii)

In triangle POQ

$O+r_1+r_2={180}^{\circ }$ ...(iii)

Comparing equations (ii) and (iii), we have

$A=r_1+r_2$

Substituting this value in equation (i)

$\delta =i+e-A$

$\delta +A=i+e$ ...(iv)

So angle of deviation produced by a prism depends upon the angle of incidence, refracting angle of prism, and the material of the prism.



When prism is in the position of minimum deviation,

then i=e and $r_1=r_2$

From equation (iv),

${\delta }_m+A=i+i$

$i=\frac{({\delta }_m+A)}{2}$

(b) $\mu =\frac{sin\left(\frac{A+{\delta }_{min}}{2}\right)}{sin\frac{A}{2}}$

But $A={\delta }_m$

$\therefore \mu =\frac{sin\left(\frac{A+A}{2}\right)}{sin\frac{A}{2}}$

$=\frac{sinA}{sin\frac{A}{2}}=\frac{2sin\frac{A}{2}.cos\frac{A}{2}}{sin\frac{A}{2}}=2cos\frac{A}{2}$

If $A={60}^{\circ }$

$\mu =2cos\frac{{60}^{\circ }}{2}=2cos{30}^{\circ }$

$=2\times \frac{\sqrt{3}}{2}=\sqrt{3}$