Question

Convex surface of the thin concavo-convex lens of refractive index 1.5 is silvered as shown. A small object kept in air at 30 cm left of the lens on its principal axis. The distance of the final image from mirror is

A. 20 cm
B. 30 cm
C. 10 cm
D. 15 cm

Answer 1

Hint:Here the refractive index of the thin concavo-convex lens is given and it is given that the convex surface of the lens is silvered. An object is kept at 30 cm left of the lens on its principal axis, we have to find the distance of the final image from the mirror. By applying lens makers formula and mirror formula, we can find the distance of the final image.

Formula used:
Lens makers formula:
$\dfrac{1}{f}=(\mu -1)\left( \dfrac{1}{{{r}_{1}}}-\dfrac{1}{{{r}_{2}}} \right)$
Where f= focal length, μ=refractive index, r1, r2=radius of curvature of both sides.
Mirror formula:
$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$
Where f=focal length, v=image distance and u=object distance.

Complete step by step solution:
Thin concavo-convex lens has refractive index 1.5. Its convex side is coated with silver. That is, it is now acting as a mirror not lens. Given a small object is kept in air at 30 cm left of the lens on its principal axis as shown in the figure.

We have to find the distance of the final image from the mirror.
On applying Lens makers formula, we get focal length of lens as:
We have equation as:
$\dfrac{1}{f}=(\mu -1)\left( \dfrac{1}{{{r}_{1}}}-\dfrac{1}{{{r}_{2}}} \right)$
On substituting values, we get:
$\dfrac{1}{f}=(1.5-1)\left( -\dfrac{1}{60}+\dfrac{1}{20} \right)$
On solving,
Focal length of lens, $f=60cm$

But we know that the convex side of a thin concavo-convex lens is silvered so it acts as a mirror. Now focal length of combination of lens and mirror is
$\dfrac{1}{{{f}_{eq}}}=\dfrac{1}{{{f}_{mirror}}}-\dfrac{2}{f}=-\dfrac{1}{10}-\dfrac{2}{60}=\dfrac{-2}{15}cm$
$\therefore {{f}_{eq}}=\dfrac{-15}{2}$
Now applying a mirror formula to find the distance of the final image.
That is,
$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=-\dfrac{2}{15}+\dfrac{1}{30}=-\dfrac{1}{10}cm$
That is, the distance of the final image from the mirror is 10 cm.

Therefore, the answer is option C.

Notes: In this kind of problem do not forget to apply sign convention. Here when we solve a problem, we use sign convention and we ignore the sign on the final answer because the sign says the position of the image with respect to the lens or mirror we are using and magnitude only matters.