Question

Consider the situation described in the previous problem. Where should a point source be placed on the principal axis so that the two images form at the same place?

Answer 1

Given,
Convex lens of focal length (f_l) = 15 cm
Concave mirror of focal length (f_m) = 10 cm
Distance between lens and mirror = 50 cm
Thus, two images will be formed,
(a) One due to direct transmission of light through lens.
(b) One due to reflection and then transmission of the rays through lens.

Let the point source be placed at a distance of 'x' from the lens as shown in the Figure, so that images formed by lens and mirror coincide.
For lens,
We use lens formula: 
$$\frac{1}{v}-\frac{1}{u}=\frac{1}{15}$$
$$\Rightarrow v=\frac{15x}{x-15}...(i)$$
For mirror,
Object distance will be (50 − x)
We use the formula:
$$\frac{1}{v}+\frac{1}{u}=\frac{1}{15}$$ 
u = − (50 − x)
f_m = − 10 cm
$$So,\frac{1}{v_m}+\frac{1}{-(50-x)}=-\frac{1}{10}$$
$$\Rightarrow \frac{1}{v_m}=\frac{1}{50-x}-\frac{1}{10}$$
$$=\frac{10-50+x}{10(50-x)}$$
$$v_m=\frac{x-40}{10(50-x)}...(ii)$$
Since the distance between lens and mirror is 50 cm,
v − v_m = 50
from equation (i) and (ii):
$$\Rightarrow \frac{15x}{x-15}-\frac{10(50-x)}{(x-40)}=50$$ 
⇒(3x² − 120x) − (100x − 2x² − 1500 + 30x)
                                 = 10 (x² − 55x + 600)
⇒ 5x² − 250x − 1500 = 10x² − 550x + 6000
⇒ 5x² − 300x + 4500 = 0
⇒       x² − 60x + 900 = 0
⇒                  (x − 30)² = 0
x = 30 cm.
∴ Point source should be placed at a distance of 30 cm from the lens on the principal axis, so that the two images form at the same place.