Question

Draw a diagram to show the formation of image of an object placed between the pole and focus of a concave mirror, derive a formula connecting objective distance, image distance and focal length for this particular case for the given concave mirror.

Answer 1

When an object is placed between the pole and the focus of a concave mirror, erect and enlarge image is formed behind the mirror.

Now from the triangles, NLF and A'B'F are similar.

So, $\frac{A^{\prime }{B}^{\prime }}{NL}=\frac{A^{\prime }F}{NF}------\left(1\right)$

Since the aperture of the concave mirror is small. So point N lies very close to pole P.

Therefore NF=PF and NL= AB------ (2)

Also, triangles ABC and A'B'C are similar.

Therefore, $\frac{A^{\prime }{B}^{\prime }}{AB}=\frac{A^{\prime }C}{AC}=\frac{P{A}^{\prime }+PC}{PC-PA}-------\left(3\right)$

From (2) and (3) $\frac{P{A}^{\prime }+PF}{PF}=\frac{P{A}^{\prime }+PC}{PC-PA}------\left(4\right)$

Applying sign convention, PF=f, PC=R=2f

(Since R=2f, PA= u)

Therefore, equation (4) becomes $\frac{-v+f}{f}=\frac{-v-2f}{2f-u}$

or -2vf+uv+$2f^2-uf=-vf+2f^2$

or uv=uf+vf

Deviding the above equation by uvf, we get,

$\frac{uv}{uvf}=\frac{uf}{uvf}+\frac{vf}{uvf}$

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