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 ( $\mu = 4 / 3$ ) at a distance of 200 cm from the refraction surface. The position of the image formed is :

• A. 120 cm
• B. 240 cm
• C. 100 cm
• D. 60 cm

Answer 1

Answer: (B)

240 cm

Answer 2

Using, $- \frac { \mu _ { 1 } } { u } + \frac { \mu _ { 2 } } { \mathrm { v } } = \frac { \mu _ { 2 } - \mu _ { 1 } } { \mathrm { R } }$

Here, $\mathrm { R } = 20 \mathrm {~cm} , \mu _ { 1 } = \frac { 4 } { 3 } , \mu _ { 2 } = 1.60$

$\therefore - \frac { 4 / 3 } { - 200 } + \frac { 1.60 } { \mathrm { v } } = \frac { 1.60 - ( 4 / 3 ) } { 20 }$