Question

An optical arrangement consists of two concave mirrors \( M_{1} \) and \( M_{2} \) and a convex lens \( L \) with a common principal axis, as shown in the figure. The focal length of \( L \) is 10 cm . The radii of curvature of \( M_{1} \) and \( M_{2} \) are 20 cm and 24 cm , respectively. The distance between \( L \) and \( M_{2} \) is 20 cm . A point object \( S \) is placed at the mid-point between \( L \) and \( M_{2} \) on the axis. When the distance between \( L \) and \( M_{1} \) is \( n / 7 cm \), one of the images coincides with \( S \). The value of \( n \) is [JEE (Adv)-2023]

Answer 1

Case-1 : If 1^st refraction on lens:

Given that the object is positioned at the focal point, the resulting light rays will transition to a parallel alignment.

Upon the first reflection at Mirror M₁:

As the incoming light rays are parallel, the resultant image will be formed at the focal point.

This v will be object for M₂, and image should be at

Case-2 : If 1^st  reflection on mirror M₂:

For M₂

\(\frac1{V_1} + \frac1{-10} = \frac1{-12}\)

V₁ = 60 cm

Then refraction on lens L

u₂ = 80 cm

\(\frac1{V_2} + \frac1{-60} = \frac1{10}\)

\(V_2= \frac{10}7\)

Then reflection on m₂

Either V₂ is at centre (normal incidence)

\(d - \frac{80}7 = 20\)

\(d = \frac {220}7\)

\(\frac n7 = \frac{220}{7}\)

n = 200

 

V₂ is at pole of m₂

\(d - \frac{80}7 = 0\)

\(d = \frac{80}7\)

\(\frac n7 =\frac{80}7\)

\(n = 80\)