Question

A converging mirror M₁, a point source S and a diverging mirror M₂ are arranged as shown in figure. The source is placed at a distance of 30 cm from M₁. The focal length of each of the mirrors is 20 cm. Consider only the images formed by a maximum of two reflections. It is found that one image is formed on the source itself. (a) Find the distance between the two mirrors. (b) Find the location of the image formed by the single reflection from M₂.

Answer 1

Given,
Converging mirror M₁ with focal length (f₁) = 20 cm
Converging mirror M₂ with focal length (f₂) = 20 cm
f₁ = f₂ = 20 cm = f
Point source is at a distance of 30 cm from M₁.


As the 1^st reflection is through the mirror M₁,
u = −30 cm
f = −20 cm

Using mirror equation:
\[\Rightarrow   \frac{1}{v} + \frac{1}{u} = \frac{1}{f}\]
\[\Rightarrow   \frac{1}{v} + \frac{1}{- 30} =  - \frac{1}{20}\] 

\[\Rightarrow   \frac{1}{v} =  - \frac{1}{20} + \frac{1}{30}\] 

\[ \Rightarrow   \frac{1}{v} =  - \frac{1}{60}\] 

\[ \Rightarrow v =  - 60  \text{ cm }\]
and for the 2^nd reflection at mirror M₂,
u = 60 − (30 + x) = 30 − x
v = − x, f = 20 cm
Again using the mirror equation:
\[\Rightarrow   \frac{1}{30 - x} - \frac{1}{x} = \frac{1}{20}\]
\[\Rightarrow   \frac{x - 30 + x}{x(30 - x)} = \frac{1}{20}\]
⇒ 40x − 600 = 30x − x²
⇒ x² + 10x − 600 = 0
\[\Rightarrow   x = \frac{10 + 50}{2} = \frac{40}{2}\]
= 20 cm or − 30 cm
∴ Total distance between the two lines is 20 + 30 = 50 cm
(b) Location of the image formed by the single reflection from M₂ = 60 \[-\] 50 = 10
Thus, the image formed by the single reflection from M₂ is at a distance of 10 cm from mirror M_2.