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Geometrical Optics

Section: Physics

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Question : 8 of 56

Marks: +1, -0

An optical arrangement consists of two concave mirrors

M_1

M_2

, and a convex lens

L

with a common principal axis, as shown in the figure. The focal length of

L

10\ \mathrm{cm}

. The radii of curvature of

M_1

M_2

20\ \mathrm{cm}

24\ \mathrm{cm}

, respectively. The distance between

L

M_2

20\ \mathrm{cm}

. A point object

S

is placed at the mid-point between

L

M_2

on the axis. When the distance between

L

M_1

\frac{n}{7}\ \mathrm{cm}

, one of the images coincides with

S

n

[JEE Adv 2023 P1]

80
150
200
220

Solution:

For reflection from

M_2

\frac{1}{v}+\frac{1}{(-10)}=\frac{1}{(-12)}

\frac{1}{v}=\frac{1}{10}-\frac{1}{12}

v=+60\ \mathrm{cm}\ (\text{ for } I_1)

For refraction from

L

\frac{1}{v}-\frac{1}{u}=\frac{1}{f}

\frac{1}{v}-\frac{1}{(-80)}=\frac{1}{10}

v=+\frac{80}{7}\ (\text{ For } I_2)

This image should be at focus of

M_1

\therefore \frac{20}{2}+\frac{80}{7}=\frac{n}{7}

n=150

Also,

If

I_2

is formed at pole of

M_1

then

\frac{n}{7}=\frac{80}{7}

n=80

And further if

I_2

is formed at centre of curvature of

M_1

then

\frac{n}{7}=\frac{80}{7}+20

\therefore n=220

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