A convex lens of focal length `15 cm` is split into two halves and the two halves are placed at a separation of `120 cm.` Between, the two halves of convex lens a plane mirror is placed horizontal and at a distance of `4 mm` below the principal axis of the lens halves. An object of length `2 mm` is placed at a distance of `20 cm` from one half lens as shown in figure.

Find the position and size of the final image.
Trace the path of rays forming the image.

Answer

A, B, C, D

Explanation

For refraction at first half lens `(1/v-1/u=1/f)`
`1/v-1/(-20)=1/15`
`:. V=60 cm`
Magnification, `m=v/u=60/(-20)=-3`
The image formed by first half lens is shown if
figure .
`AB=2mm, A_1B_1=6mm, AO_1=20 cm,`
`O_1F=15 cm` and `O_1A_1=60 cm`

Point `B_1` is `6 mm` below the principal axis of the
lenses. Plane mirror is 4mm below it.

Hence, `4mm` lenght of `A_1B_1(i.e.A_1C_1)` acts as
real object for mirror. Mirror forms its virtual
image `A_2C_2. 2mm` length of `A_1B_1 (i.e.C_1B_1)` acts
as virtual object for mirror. Real image `C_2B_2` is
formed of this part. Image formed by plane mirror
is shown in figure (b) .
For the second half lens, `1/v-1/(-60)=1/15`
`:. v=+20`
`m=v/u=20/(-60)=-1/3`
So, length of final image
`A_3B_3=1/3A_2B_2=2mm`
Point `B_2` is 2mm below the optic axis of second half
lens. Hence, its image `B_3` is formed `2//3 mm` above
the principal axis. Somilarly, point `A_2` is 8mm
below the principal axis. Hence, its image is `8//3 mm`
above it. Therefore, image is at a distance of 20 cm
behind the second half and at a distance of `2//3 mm`
above the principal axis. The size of image is 2 mm
and is inverted as compared to the given object.
Image formed by second half lens is shown in
figure

Ray diagram for final image is shown in
figure