Question:

A thin convex lens L (refractive index = 1.5) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18cm, 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?

4/3

3/2

Correct option is B. 4/3

1/f₁ = (1.5 - 1)/8 = 1/18

When μ₁ is filled between lens and mirror
1/f₂ = (μ₁ - 1)(-1/8)

P = 2/18 - 2/18(μ₁ - 1) = 2 - 2μ₁ + 2/18 = 20 - 2μ₁/18

F = -(18/2 - μ₁) / 2 = (6 - 3μ₁)/3
3μ₁ = 4
μ₁ = 4/3