Question

In the given figure, the radius of curvature of curved surface for both the plano-convex and plano-concave lens is 10 cm and refractive index for both is 1.5. The location of the final image after all the refractions through lenses is

Answer 1

The correct option is B 20 cm

The focal length of the plano - convex lens is:

$\frac{1}{f_1}=(1.5-1)(\frac{1}{+10}-\frac{1}{\infty })=\frac{1}{20}$

Focal length of plano - concave lens is:

$\frac{1}{f_2}=(1.5-1)(\frac{1}{\infty }-\frac{1}{10})=-\frac{1}{20}$

since parallel beams are incident on the lens, its image from plano - concave lens will be formed at $+20cm$ from it (at the focus) and will act as an object for the plano - concave lens. since the two lens are at a distance of $10cm$ from each other, therefore, for the next lens

$u=+10cm$

$v=\frac{uf}{u+f}=\frac{10\times 20}{10-20}$

$\Rightarrow v=\left(\frac{200}{10}\right)cm=20cm$
Hence, the correct option is $\left(B\right)$