Behind a thin converging lens having both the surfaces of the same radius 10 cm, a plane mirror has been placed. The image of an object at a distance of 40 cm from the lens is formed at the same position. What is the refractive index of the lens?

Question

Behind a thin converging lens having both the surfaces of the same radius 10 cm, a plane mirror has been placed. The image of an object at a distance of 40 cm from the lens is formed at the same position. What is the refractive index of the lens? (A) 1.5 (B) 5 / 3 (C) 9 / 8 (D) None of these

Tardigrade Admin · Accepted Answer

" O " act as the focal point. Also, (1/f)=(μ-1)((1/R1)-(1/R2)) ⇒ (1/40)=(μ-1)((1/10)+(1/10)) (1/40)=(μ-1)((2/10)) ⇒ μ=9 / 8

Q. Behind a thin converging lens having both the surfaces of the same radius $10 c m$ , a plane mirror has been placed. The image of an object at a distance of $40 c m$ from the lens is formed at the same position. What is the refractive index of the lens?

$1.5$

$5/3$

$9/8$

None of these

" $O$ " act as the focal point.
Also, $\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\Rightarrow \frac{1}{40} = (μ - 1) (\frac{1}{10} + \frac{1}{10})$
$\frac{1}{40} = (μ - 1) (\frac{2}{10})$
$\Rightarrow μ = 9/8$

Q. Behind a thin converging lens having both the surfaces of the same radius 10cm, a plane mirror has been placed. The image of an object at a distance of 40cm from the lens is formed at the same position. What is the refractive index of the lens?

1.5

5/3

9/8