Question

A bi-convex lens is formed with two plano-convex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both curved surfaces are of the same radius of curvature R = 14 cm. For this bi-convex lens, for an object distance of 40 cm, the image distance will be

• A. $-280.0cm$
• B. $40.0cm$
• C. $21.5cm$
• D. $13.3cm$

Answer 1

The correct option is B $40.0cm$

We consider refraction from the three spherical surfaces,

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

For refraction from first surface,

$\frac{1.5}{v_1}-\frac{1}{-40}=\frac{1.5-1}{+14}$

For refraction from second surface,

$\frac{1.2}{v_2}-\frac{1.5}{v_1}=\frac{1.2-1.5}{\infty }$

For refraction from third surface,

$\frac{1}{v_3}-\frac{1.2}{v_2}=\frac{1-1.2}{-14}$

Adding the three equations gives,

$\frac{1}{v_3}=\frac{-1}{40}+\frac{1}{20}=\frac{1}{40}$

Hence, $v_3=40cm$