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

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

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 $+20\, cm$ 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 $10 \,cm$ from each other, therefore, for the next lens
$u = +10 \,cm$.
$\therefore v= \frac{uf}{u + f}$
$ =\frac{10\times 20}{10 -20} = 20\,cm$