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 refracting 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

Using, $-\frac{{\mu }_1}{u}+\frac{{\mu }_2}{v}=\frac{{\mu }_2-{\mu }_1}{R}$
Here, R = 20 cm, ${\mu }_1=\frac{4}{3},{\mu }_2=1.60,u=-200cm$
$\therefore -\frac{4/3}{-200}+\frac{1.60}{v}=\frac{1.60-\left(4/3\right)}{20}\Rightarrow v=240m$