Question

A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of :

(1) 20 cm from the lens, it would be a real image. 

(2) 30 cm from the lens, it would be a real image. 

(3) 30 cm from the plane mirror, it would be a virtual image. 

(4) 20 cm from the plane mirror, it would be a virtual image.

Answer 1

Let us use the lens formula

\(\frac 1 v -\frac 1u = \frac 1f\)

Given

u = -60 cm

f = 30 cm

v₁ = ?

For the first refraction from a convex lens

\(\frac 1{v_1} + \frac 1{60} = \frac 1{30}\)

So we get

v₁ = 60 cm

l₁ is the first image by the lens

An image will be produced by the plane image at a 20 cm distance to the left of it

For the second refraction from a convex lens

u = -20 cm

v = ?

f = 30 cm

Using the lens formula

\(\frac 1v - \frac 1{-20} =\frac 1{30}\)

v = -60 cm

The final image is virtual and at a distance of 60 - 40 = 20 cm from the plane mirror.

Therefore, the final image would be formed at a distance of 20 cm from the plane mirror; it would be a virtual image.

Answer 2

Option : (4). 20 cm from the plane mirror, it would be a virtual image.

first, for image formation from lens 

u = – 60 cm 

f = + 30 cm 

⇒ v = \(\frac{uf}{u+f}\) = \(\frac{-60\times30}{-60+30}\)= 60cm

this real image formed by lens acts as virtual object for mirror

Real image from plane mirror is formed 20 cm in front of mirror, hence at 20 cm distance from lens. 

Now, 

for second refraction from lens, 

u = – 20 cm 

f = +30 cm

v = \(\frac{uf}{u+f}\) = \(\frac{-20\times30}{-20+30}\)= - 60cm

So, 

Final virtual image is 60 cm from lens, or 20 cm behind mirror.