Question

A convex surface of radius of curvature 40cm separates two media of refractive indices 4/3 and 1.50. An object is kept in the first medium at a distance of 20cm from the surface. Calculate the position of the image.

• A. $20cm$
• B. $24cm$
• C. $30cm$
• D. $32cm$

Answer 1

The correct option is B $24cm$



$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}{\mu }_2=\frac{3}{2};{\mu }_1=\frac{4}{3};u=-20cm;R=40cm\frac{3}{2v}+\frac{\frac{4}{3}}{20}\frac{3}{2v}+\frac{\frac{1}{3}}{\frac{20}{1}}=\frac{\frac{3}{2}-\frac{4}{3}}{40}\frac{3}{2v}+\frac{4}{60}=\frac{\frac{9-8}{6}}{40}=\frac{1}{240}\frac{3}{2v}=\frac{1}{240}-\frac{4}{60}=\frac{1-16}{240}=-\frac{15}{240}\frac{3}{2v}=-\frac{15}{240}v=\frac{3\times 240}{2\times 15}v=-24cm$