Question

Find the focal length of the lens shown in figure. The refractive index of lens material is $\frac{3}{2}$.

Answer 1

We know,
$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

Here let us take light to be incident from the left side.

$\Rightarrow$ $R_1=-R=-10\text{cm}$

$R_2=+R=+10\text{cm}$

Also, ${\mu }_2=\frac{3}{2},{\mu }_1=1$

$\frac{1}{f}=(\frac{3}{2}-1)(\frac{1}{-10}-\frac{1}{10})$

$\Rightarrow \frac{1}{f}=\frac{1}{2}\times \frac{-2}{10}$

$\therefore f=-10\text{cm}$

Why this question? Remark: Negative focal length of the lens system represents that lens shows a diverging behaviour.