Question

If an incident light ray parallel to the principal axis is incident at an angle of $i$ as shown in the figure, then find the distance of $Q$ from the centre of curvature $C$ if the angle of incidence is finite (not very small). Here, $R$ is the radius of curvature.

• A. $2R\sin i$
• B. $\frac{R}{2\cos i}$
• C. $\frac{R}{2{\sec }^{2}i}$
• D. $2R\tan i$

Answer 1

The correct option is B $\frac{R}{2\cos i}$

Here, the angle of incidence $i$ is finite (not very small).
The reflected light ray will not get converged at the focal point.

From laws of reflection,
$\angle i=\angle r$
Here, $CM$ is the radius of curvature.
From alternate angles, $\angle NCQ=\angle i$
$\because \angle NMQ=\angle NCQ$
Thus, $CQ=QM$ and $△CMQ$ is an isosceles triangle.
Drawing a perpendicular from $Q$ on $CM$ will bisect $CM$.
So, $CN=MN$
or, $CN=\frac{CM}{2}=\frac{R}{2}$ ...............$\left(1\right)$
In $△CNQ$,
$\cos i=\frac{CN}{CQ}$
$\Rightarrow CQ=\frac{CN}{\cos i}$
$\Rightarrow CQ=\frac{R}{2\cos i}$ $[$ from $\left(1\right)]$