A convex refracting surface of radius of curvature 20 cm separates two media of refractive indices 4/3 and 1.60. An object is placed in the first medium (μ = 4/3) at a distance of 200 cm from the refracting surface. The position of the image formed is

Question

A convex refracting surface of radius of curvature 20 cm separates two media of refractive indices 4/3 and 1.60. An object is placed in the first medium (μ = 4/3) at a distance of 200 cm from the refracting surface. The position of the image formed is (A) 120 cm (B) 240 cm (C) 100 cm (D) 60 cm

Tardigrade Admin · Accepted Answer

Using, -(μ1/u)+(μ2/v) = (μ2-μ1/R) Here, R = 20 cm, μ1=(4/3), μ2 = 1.60, u = -200 cm ∴ - (4/3/-200) +(1.60/v) = (1.60-(4/3)/20) ⇒ v = 240 m

Q. A convex refracting surface of radius of curvature $20 c m$ separates two media of refractive indices $4/3$ and $1.60$ . An object is placed in the first medium $(μ = 4/3)$ at a distance of $200 c m$ from the refracting surface. The position of the image formed is

$120 c m$

$240 c m$

$100 c m$

$60 c m$

Using, $- \frac{μ _{1}}{u} + \frac{μ _{2}}{v} = \frac{μ _{2} - μ _{1}}{R}$
Here, $R = 20 c m$ ,
$μ_{1} = \frac{4}{3}, μ_{2} = 1.60, u = - 200 c m$
$∴ - \frac{4/3}{- 200} + \frac{1.60}{v}$
$= \frac{1.60 - ( 4/3 )}{20}$
$\Rightarrow v = 240 m$

Q. A convex refracting surface of radius of curvature 20cm separates two media of refractive indices 4/3 and 1.60. An object is placed in the first medium (μ=4/3) at a distance of 200cm from the refracting surface. The position of the image formed is

120cm

240cm

100cm

60cm