Question

(a) Draw a ray diagram to show image formation when the concave mirror produces a real, inverted and magnified image of the object.
(b) Obtain the mirror formula and write the expression for the linear magnification.
(c) Explain two advantages of a reflecting telescope over a refracting telescope.

Answer 1

(a) Image produced by a concave mirror is real, inverted and magnified for an object placed between F and C :


(b) Derivation for mirror formula and magnification :
Consider an object AB be placed in front of a concave mirror beyond centre of curvature C.


If F is the focus, focal length = f, object distance = u, image distance = v

As $△$ s ABC and A'B'C are similar,

$\frac{AB}{A^{\prime }{B}^{\prime }}=\frac{CB}{C{B}^{\prime }}$ ...(i)

Again, as $△$ s ABP and A'B'P are similar

$\frac{AB}{A^{\prime }{B}^{\prime }}=\frac{PB}{B^{\prime }P}$ ...(ii)

$\Rightarrow \frac{CB}{C{B}^{\prime }}=\frac{PB}{P{B}^{\prime }}$ ...(iii)

$\therefore$ From (iii), $\frac{PB-PC}{PC-P{B}^{\prime }}=\frac{PB}{P{B}^{\prime }}$

$\Rightarrow \frac{-u+R}{-R+V}=\frac{-u}{v}[\because$ PB=-u, PC =-R PB' =-v]

$\Rightarrow$ uR+vR= 2uv

$\Rightarrow \frac{1}{v}+\frac{1}{u}=\frac{2}{R}$
($\because$ R= 2f)

Linear magnification, m = $\frac{\text{Image height}\left(h_1\right)}{\text{Object height}\left(h_0\right)}$

From equation (ii), $\frac{A^{\prime }{B}^{\prime }}{AB}=\frac{P{B}^{\prime }}{PB}$

Using new Cartesian sign conventions, A'B' =$-h_{2}AB=+h_1$, PB' =-v, PB=-u

$\therefore \frac{-h_2}{h_1}=\frac{-v}{u}=\frac{v}{u}$
$m=\frac{h_2}{h_1}=\frac{-v}{u}$
(c) Advantages of reflecting telescope over a refracting telescope are :
(i) Reflecting telescope is free from chromatic and spherical aberrations unlike refracting telescope. Thus image formed is sharp and bright.
(ii)It has a larger light gathering power so that a bright image of even far off object is obtained.