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प्रश्न
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.
उत्तर
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 1st 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 2nd 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 M2.
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