Question

A bi-convex lens is formed with two thin plano convex lenses as shown in the figure. Refractive index $\mu$ 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 $R=14\text{cm}$. For an object placed at a distance of $40\text{cm}$ infront of lens, the image distance will be

Answer 1

For plano convex lens,
$f_1=\frac{R}{({\mu }_1-1)}=\frac{14}{1.5-1}=28$ cm
$f_2=\frac{R}{({\mu }_2-1)}=\frac{14}{1.2-1}=70$ cm
Equivalent focal length of setup is
$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{28}+\frac{1}{70}$
$\frac{1}{f_{eq}}=\frac{5+2}{140}=\frac{7}{140}=\frac{1}{20}$
$f_{eq}=20$ cm
Now, using lens formula,
$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}-\frac{1}{40}$
$\frac{1}{v}=\frac{2-1}{40}=\frac{1}{40}$
$v=40$ cm