Question

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

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

Answer 1

Answer: (B)

$\frac{1}{f_{1}}=\left(1 . 2 - 1\right)\left(\frac{1}{\infty } - \frac{1}{- 14}\right)=\frac{0 . 2}{14}$
$\frac{1}{f_{2}}=\left(1 . 5 - 1\right)\left(\frac{1}{14} - \frac{1}{\infty }\right)=\frac{0 . 5}{14}$
$\frac{1}{f_{eq}}=\frac{0 . 7}{14}=\frac{1}{20}$
$\frac{1}{v}-\frac{1}{- 40}=\frac{1}{20}$
$v=40cm$ .