Question

Diameter or aperture of a plano-convex lens is $6 \, cm$ and its thickness at the centre is $3\,mm.$ The image of an object formed is real and twice the size of the object. If the speed of light in the material of the lens is $2\times 10^{8} \, \text{m s}^{- 1}$ . The distance where the object is placed from the plano-convex lens is $\ldots \ldots \ldots \times 15 \, cm$ .

Answer 1

$\left(R - y\right)^{2}+r^{2}=R^{2}$
$R=\frac{r^{2}}{2 y}=\frac{\left(\frac{6}{2}\right)^{2}}{2 \times 0 .3}=15 \, cm$
$\frac{1}{f}=\frac{\mu - 1}{R}$
$\mu =\frac{C}{V}=\frac{3 .0 \times 10^{8}}{2 .0 \times 10^{8}}=1.5$
$\therefore f=30 \, cm$ converging
Now
$f=30 \, cm$
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{30}=\frac{1}{v}-\frac{1}{u}$
$m=\frac{v}{u}=-2$
$v=-2u$
$\frac{1}{30}=\frac{- 1}{2 u}-\frac{1}{u}$
$\frac{1}{30}=-\frac{3}{2 u}$
$u=-45 \, cm$