Question

A point object is placed at a distance $x_1$ behind the principal focus, on the principal axis of a concave mirror. The image is formed at a distance $x_2$ behind the principal focus.The focal length of the mirror is:

• A. $x_1{x}_2$
• B. $\frac{x_1+x_2}{2}$
• C. $\frac{x_1}{x_2}$
• D. $\sqrt{x_1{x}_2}$

Answer 1

The correct option is D $\sqrt{x_1{x}_2}$

You know the mirror formula, $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$, in terms of object and image distances from the pole. Here the distances are given from the focus,ie $x_1$ and $x_2$. Let's consider the diagram below:



The distance of the object from the pole = $x_1+f$.

Therefore, applying sign convention, $u=-(x_1+f)$.

Similarly, distance of the image from the pole = $x_2+f$.

Therefore, applying sign convention, $v=-(x_2+f)$.

Now,

Substituting in the mirror formula:

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

we get,

$\frac{1}{-(x_1+f)}+\frac{1}{-(x_2+f)}=-\frac{1}{f}$

or,$\frac{(x_2+f)+(x_1+f)}{(x_1+f)(x_2+f)}=\frac{1}{f}$

or,$f(x_2+x_1+2f)=x_1{x}_2+x_{1}f+f{x}_2+f^2$

Therefore, $f^2=x_1{x}_2$

$f=\sqrt{x_1{x}_2}$

Now, note that even though we did this for an object placed beyond the focus, the same relationship can be obtained for a general object, and also for a convex mirror. In fact, this is a form of expressing the mirror formula, due to Newton!