Question

What is the minimum value of $d$ for which final image and object are at the same place

A. $30\,cm$
B. $40\,cm$
C. $20\,cm$
D. $80\,cm$

Answer 1

An optical lens is made up of two spherical surfaces in most cases. The lens is called a biconvex lens or simply convex lens if certain surfaces are bent outwards. A concave lens is one that has at least one inwardly curving surface.

Complete step by step answer:
In this question, we can see two lenses. That is, convex lens and concave in the first and second places, respectively. Because of two lenses, two images will form here. The first image will be formed by the convex lens, which is the first lens. Take the image to be formed at a distance ${v_1}$ when the object is at a point $O$ at the distance ${u_1} = - 30\,cm$
It is given the focal length of the convex lens, ${f_1} = 20\,cm$.
The image formed by the convex lens will act as the object for the concave lens.
That is, ${v_1} = {u_2}$
It is given the focal length of the concave lens, ${f_2} = - 10\,cm$.

Now we are using the lens formula for the convex lens.
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
We are substituting the given values for the convex lens in the formula to find ${v_1}$ .
That is,
$\dfrac{1}{{{v_1}}} - \dfrac{1}{{ - 30}} = \dfrac{1}{{20}}$
$\Rightarrow \dfrac{1}{{{v_1}}} + \dfrac{1}{{30}} = \dfrac{1}{{20}}$
$\Rightarrow \dfrac{1}{{{v_1}}} = \dfrac{1}{{20}} - \dfrac{1}{{30}}$
$\Rightarrow \dfrac{1}{{{v_1}}} = \dfrac{{30 - 20}}{{20 \times 30}}$
$\Rightarrow \dfrac{1}{{{v_1}}} = \dfrac{{10}}{{600}} = \dfrac{1}{{60}}$
Hence, ${v_1} = 60$
We want the final image and object at the same place. That is,
$\therefore d = 60 - 20 = 40\,cm$

So, the answer is option B.

Additional Information:
We can use the lens formula to find out ${v_2}$ .
$\dfrac{1}{{{v_2}}} - \dfrac{1}{{60}} = \dfrac{1}{{ - 10}}$
$\Rightarrow \dfrac{1}{{{v_2}}} = \dfrac{1}{{ - 10}} + \dfrac{1}{{60}}$
$\Rightarrow \dfrac{1}{{{v_2}}} = \dfrac{{60 - 10}}{{60 \times 10}}$
$\Rightarrow \dfrac{1}{{{v_2}}} = \dfrac{{50}}{{600}} = \dfrac{5}{{60}}$
Hence, ${v_2} = \dfrac{{60}}{5} = 12cm$
So, we found that the final image formed at $12cm$ .

Note: Remember the formula $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$. Convex mirrors have a reflecting surface that bulges outwards, whereas concave mirrors have a reflecting surface that bulges inwards. The images that form in these two mirrors are the most significant distinction. In other words, convex mirrors produce smaller images, and concave mirrors produce larger images.