Question

A plano-convex of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens, an object be placed in order to have a real image of the size of the object?

Answer 1

From lens makers formula, $\frac{1}{f_l}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})=(1.5-1)(\frac{1}{\infty }-\frac{1}{-30})=\frac{1}{60}$

Also, the focal length of the combination can be given by :

$-\frac{1}{F}=\frac{2}{f_l}-\frac{1}{f_m}$

$-\frac{1}{F}=\frac{2}{60}-\frac{1}{-15}$

$\Rightarrow F=-10cm$

Hence the object should be placed at a distance $2F=2\times 10cm=20cm$ from lens to get a real image of the same size of the object.