JEE Advanced 2020 Paper 1 · Physics · Ray and Wave Optics · hard

A point object lies on the principal axis of a thin convex lens of focal length 20 cm, at a distance of 30 cm to the left of the lens. A plane mirror…

Numerical answer

A point object lies on the principal axis of a thin convex lens of focal length

20\ \mathrm{cm}

, at a distance of

30\ \mathrm{cm}

to the left of the lens. A plane mirror is placed perpendicular to the axis at

48\ \mathrm{cm}

to the right of the lens. Light from the object passes through the lens, gets reflected by the mirror and passes again through the lens. A small object of height

h

perpendicular to the axis is considered. The magnitude of the overall lateral magnification of the final image, relative to the object, is (give the answer correct to two decimal places)

Show answer and solution

Answer: 2.50

First refraction:

\frac1v-\frac1{-30}=\frac1{20}

gives

v_1=60

cm to the right, with

m_1=\frac{v}{u}=-2

. The mirror, at 48 cm, intercepts the converging rays before they meet, so it acts on a virtual object

12

cm behind it. A plane mirror forms a real image

12

cm in front of the mirror, i.e.

48-12=36

cm from the lens, with magnification

+1

. This real image is a real object for the second pass of the lens,

u=-36

cm (distances now measured in the direction of the travelling light):

\frac1v=\frac1{20}-\frac1{36}=\frac{16}{720}

, giving

v_2=45

cm and

m_2=\frac{v}{u}=-1.25

. The overall magnification is

(-2)(+1)(-1.25)=+2.5

, so its magnitude is

2.50

and the image is erect.

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Original question written in the JEE Advanced 2020 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake