Question

A pin is placed $10cm$ in front of a convex lens of focal length $20cm$ and refractive index $1.5$.The surface of the lens further away from the pin is silvered. The final image from the lens is at a distance of :
A.$10cm$
B.$20cm$
C.$12cm$
D.$13cm$

Answer 1

Hint: Here we will use Lens Maker’s formula to solve the problem. The surface of the lens is silvered then the lens will behave as a concave mirror. Form the ray diagram to get the vision of the solution.
Formula used:
\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]; where $v$is the final position of the image,$u$ is the position where the object is placed in front of lens and $f$is the focal length of the lens.

Complete answer:
From the given question we have the following information:
$u = 10cm$(the distance of the pin from the lens)
$f = 20cm$(focal length of the convex lens)
To solve this problem, first we need to correctly assign the signs to the both sides of the lens. Then we are going to use the Lens Maker’s formula for the solution :
\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]; this equation is also known as Gaussian form of lens equation.
We already have the values of $u$and $f$from the question. So, substituting them in the Lens Maker’s formula we get:
$ \Rightarrow \dfrac{1}{v} - \left( {\dfrac{1}{{ - 10}}} \right) = \dfrac{1}{{20}}$
$ \Rightarrow \dfrac{1}{v} = \left( {\dfrac{1}{{20}} - \dfrac{1}{{10}}} \right)cm$
$ \Rightarrow \dfrac{1}{v} = \dfrac{{1 - 2}}{{20}}cm$
$ \Rightarrow \dfrac{1}{v} = \dfrac{{ - 1}}{{20}}cm$
$ \Rightarrow v = - 20cm$
Thus, the final image of the lens is formed at a distance of 20 cm from the lens on the same side of the lens where the pin was placed.

Hence, the correct option is B.

Additional information:
Hence image after first reflection is formed on the center of curvature of mirror. Hence it would retrace the path. Thus, the final image would obtained at the same position

Note:
The signs have to be chosen correctly otherwise the position of the image would get misrepresented. The image will be formed in front of the concave mirror and it is the real image. Memorizing ray optics diagram can help you to solve these kind of problems