Question

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

• A. $15cm$
• B. $20cm$
• C. $25cm$
• D. $40cm$

Answer 1

Answer: (B)

The focal length of the plano-convex lens is
$\frac{1}{f}=\left(1.5-1\right)\left(\frac{1}{+10}-\frac{1}{\infty }\right)=\frac{1}{20}$
Focal length of plano-concave lens is
$\frac{1}{f}=\left(1.5-1\right)\left(\frac{1}{\infty }-\frac{1}{10}\right)=\frac{-1}{20}$
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
$u=+10cm$.
$\therefore v=\frac{uf}{u+f}$
$=\frac{10\times 20}{10-20}=20cm$