Question

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

Answer 1

Given,
${\mu }_1=1.5$

${\mu }_2=1.2$

$R_1=R_2=R=14cm$

$u=40cm$

The equivalent focal length by the combination of lenses is
$\{\because \text{By the lens maker formula}\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})\}$

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$

$({\mu }_1-1)(\frac{1}{R}-\frac{1}{\infty })+({\mu }_2-1)(\frac{1}{\infty }-\frac{1}{(-R)})$

$\Rightarrow f=20cm$

By the lens formula is
Now $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\Rightarrow v=-40cm$