Question

Diameter or aperture of a plano-convex lens is $6cm$ and its thickness at the centre is $3mm.$ 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 $\dots \dots \dots \times 15cm$ .

Answer 1

Answer: 3

${\left(R-y\right)}^2+r^2=R^2$
$R=\frac{r^2}{2y}=\frac{{\left(\frac{6}{2}\right)}^2}{2\times 0.3}=15cm$
$\frac{1}{f}=\frac{\mu -1}{R}$
$\mu =\frac{C}{V}=\frac{3.0\times 10^8}{2.0\times 10^8}=1.5$
$\therefore f=30cm$ converging
Now
$f=30cm$
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{30}=\frac{1}{v}-\frac{1}{u}$
$m=\frac{v}{u}=-2$
$v=-2u$
$\frac{1}{30}=\frac{-1}{2u}-\frac{1}{u}$
$\frac{1}{30}=-\frac{3}{2u}$
$u=-45cm$