Question

A bi-convex lens is formed with two thin 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 $\mathrm{1.2.}$ Both the curved surface 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

The correct option is B $40.0cm$

Using lens formula,

$\Rightarrow \frac{1}{v}-\frac{1}{u}=\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$

$\Rightarrow \frac{1}{v}=\frac{1}{u}+\frac{1}{f_1}+\frac{1}{f_2}$

$\Rightarrow \frac{1}{v}=\frac{1}{-40}+(1.5-1)(\frac{1}{14}-\frac{1}{\infty })+(1.2-1)(\frac{1}{\infty }-\frac{1}{-14})$

$\Rightarrow v=+40cm.$

Hence, the answer is $40.0cm.$