Question

A thin convex lens L (refractive index = 15) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index μ₁
is put between the lens and the mirror, the pin has to be moved to A’. such that OA’ = 27cm, to get its inverted real image at A’
itself. The value of μ₁ will be:
(A) √2 (B) 4/3
(C) √3 (D) 3/2

Answer 1

Hey mate

maybe your answer will be c)

Answer 2

•According to the question

A thin convex lens L (refractive index = 15) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index μ₁

is put between the lens and the mirror, the pin has to be moved to A’. such that OA’ = 27cm, to get its inverted real image at A’ itself. Now

we can say that

1/f1 = 1/2 × 2/18 = 1/18

and 1/f2 = (μ1 - 1)/(-18)

and as μ1 is filled between mirror and lens

P = 2/18 - 2/18(μ1 - 1)

= (2-2μ +2)/18

= Fn =27= -(18/2-μ1) [as OA′=27]

or,2=( 6- 3μ1)

so, μ1 = 4/3

So the answer is (B) 3/2