First slide

Ray optics

Question

The distance of a real object from the focus of a convex mirror of radius of curvature ‘a’ is ‘b’. Then the distance of the image from the focus is

Difficult

A

$\frac{b^{2}}{4 a}$
B

$\frac{a}{b^{2}}$
C

$\frac{a^{2}}{4 b}$
D

$\frac{b}{a}$

Solution

The focal length of convex mirror is f = a/2
The real object distance u = b - a/2
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \frac{1}{v} - \frac{1}{b - (a / 2)} = \frac{2}{a}$
solving we get
$v = \frac{a (2 b - a)}{4 b}$
distance of image from focus is
$\frac{a}{2} - v = \frac{a^{2}}{4 b}$

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