Question

In the given figure a plano-concave lens is placed on a paper on which flower is drawn. How far above its actual position does the flower appear to be ?

• A. 10 cm
• B. 15 cm
• C. 50 cm
• D. none of these

Answer 1

The correct option is A 10 cm

Given: In the given figure a plano-concave lens is placed on a paper on which flower is drawn.

To find how far above its actual position does the flower appear

Solution:

Refraction aoccur at the surface,

and according to the given criteria,

object distance, $u=-20cm$

refractive index, ${\mu }_1=\frac{3}{2},{\mu }_2=1$

And radius of curvature, $R=+20cm$

Applying the lens formula, we get

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}⟹\frac{1}{v}-\frac{\frac{3}{2}}{(-20)}=\frac{1-\frac{3}{2}}{20}⟹\frac{1}{v}=\frac{3}{-40}+\frac{2-3}{40}⟹\frac{1}{v}=\frac{-3-1}{40}⟹v=-10cm$

Hence the flower will appear 10 cm above its actual position of flower.