Question

A biconvex lens has radii of curvature of two faces 0.1 m and 0.12 respectively. If the focal length of the lens is 0.1 m. Find (i) the refractive index of the material of lens (ii) power of lens (iii) nature and position of the image of an object placed at a distance of 6 cm from the lens.

Answer 1

$\left(i\right)$ $\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

Here, $f=0.1m,R_1=0.1m,R_2=-0.12m$

$⟹\frac{1}{0.1}=(\mu -1)(\frac{1}{0.1}+\frac{1}{0.12})$

$⟹\frac{1}{0.1}=(\mu -1)\left(\frac{6+5}{0.6}\right)$

$⟹\mu -1=10\times \frac{0.6}{11}$

$⟹\mu =\frac{17}{11}=1.54$

$\left(ii\right)$ $P=\frac{1}{P\left(m\right)}=\frac{1}{0.1}=+10D$

$\left(iii\right)$ $f=0.1m=10cm$ and $u=-6cm$

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{u}+\frac{1}{f}=\frac{-1}{6}+\frac{1}{10}$

$⟹v=-15cm$

Magnification $m=\frac{v}{u}=\frac{-15}{-6}=2.5$

Hence, the image formed is virtual, erect and 2.5 times magnified that of the object.