An equiconvex thin lens of refractive…

An equiconvex thin lens of refractive index 1.5 has radius of curvature 20 cm for each surface. One face is silvered, so the lens acts as a mirror…

One or more options correct

An equiconvex thin lens of refractive index

1.5

has radius of curvature

20\ \mathrm{cm}

for each surface. One face is silvered, so the lens acts as a mirror system. Choose the correct statement(s).

A. The system behaves as a concave mirror of focal length $10\ \mathrm{cm}$ .
B. An object placed $7.5\ \mathrm{cm}$ from the lens on the unsilvered side forms a real, inverted image of twice its size.
C. If the refractive index were $2.0$ with the same radii, the focal length of the system would be $4\ \mathrm{cm}$ .
D. The system behaves as a convex mirror.

Show answer and solution

Answer: B

For a silvered lens, light crosses the lens, reflects, and crosses again, so

\dfrac1F=\dfrac2{f_l}+\dfrac1{f_m}

. Here

\dfrac1{f_l}=(1.5-1)\dfrac{2}{0.20}=5\ \mathrm{m^{-1}}

f_l=20\ \mathrm{cm}

, and the silvered surface of radius

20\ \mathrm{cm}

has

f_m=10\ \mathrm{cm}

. Thus

\dfrac1F=10+10=20\ \mathrm{m^{-1}}

and

F=5\ \mathrm{cm}

, a concave mirror. Statement A (

10\ \mathrm{cm}

) is false, and D is false. For

u=-7.5\ \mathrm{cm}

and

f=-5\ \mathrm{cm}

1/v=-1/5+1/7.5

gives

v=-15\ \mathrm{cm}

and

m=-v/u=-2

, a real inverted image twice as large; B is true. For

\mu=2

1/f_l=10

and

1/F=20+10=30

, so

F=3.33\ \mathrm{cm}

, not

4\ \mathrm{cm}

; C is false.

Original question written in the JEE Advanced 2016 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake