Question

A diverging lens of focal length ${{f}_{1}}$ is placed in front of and coaxially with a concave mirror of focal length ${{f}_{2}}$. Their separation is d. A parallel beam of light incident on the lens returns as a parallel beam from the arrangement. Then
(This question has multiple correct options)
A. The beam diameters of the incident and reflected beams must be the same.
$\text{B}\text{. }d=2|{{f}_{2}}|-|{{f}_{1}}|$
$\text{C}\text{. }d=|{{f}_{2}}|-|{{f}_{1}}|$
D. If the entire arrangement is immersed in water, the conditions will remain unaltered

Answer 1

Hint: If the parallel beam of light must return as a parallel beam, then the incident rays of light must trace their path after reflecting at the concave mirror. Draw a suitable diagram of the given arrangement to understand the conditions.

Formula used:
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

Complete step by step answer:
It is given that a diverging lens of focal length ${{f}_{1}}$ is placed in front of a concave mirror. Let the diverging lens be a concave lens. It is said that the concave lens is placed coaxially with the concave mirror. This means that the optical axis of the lens and the principal axis of the mirror coincide. The arrangement is shown in the given figure.
The focal length of the mirror is said to be ${{f}_{2}}$.
When the parallel beam of light passes through the concave lens, the rays of light will diverge in such a way that after passing through the lens they will appear to be emerging from focus of the lens that is in the opposite direction of the incident rays (look at the figure).

Then the diverged rays will reflect at the surface of the concave mirror and again pass through the lens. Now, if these returning rays of light have to parallel to the incident beam of light then the rays must trace the path of incidence as shown in the figure.
As you can see, the beam diameters of the incident and reflected beams must be the same.
Hence, option A is correct.
When the incident rays diverge and fall on the mirror, these rays act as the rays coming from an object placed in the left focus of the lens. The rays must reflect in such a way that they will intersect at the same focus of the lens.
Hence, use the formula of a curved mirror i.e. $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ ….. (i).
In this case, $u=-|{{f}_{1}}|-d$, $v=-|{{f}_{1}}|-d$, $f=-|{{f}_{2}}|$
Substitute the values in equation (i).
$\Rightarrow \dfrac{1}{-|{{f}_{1}}|-d}+\dfrac{1}{-|{{f}_{1}}|-d}=\dfrac{1}{-|{{f}_{2}}|}$
$\Rightarrow \dfrac{2}{|{{f}_{1}}|+d}=\dfrac{1}{|{{f}_{2}}|}$
$\Rightarrow 2|{{f}_{2}}|=|{{f}_{1}}|+d$
$\Rightarrow d=2|{{f}_{2}}|-|{{f}_{1}}|$
Hence, option B is also correct.
The position of the image of an object depends on the position of the object and the focal length of the mirror or the lens only. It does not depend on the surrounding medium if the entire set up is present in a single medium.
However, the focal length of a lens depends on the surrounding medium.
Therefore, when the entire arrangement is placed in water, the conditions will be changed.
Hence, the correct options are A and B.

Note: The position of the image of an object depends on the position of the object and the focal length of the mirror or the lens only. It does not depend on the surrounding medium if the entire set up is present in a single medium.
However, the focal length of a lens depends on the surrounding medium.
Therefore, when the entire arrangement is placed in water, the conditions will be changed.
We do not have to use the mirror formula if we know the fact that the image and the object coincide only when the object is placed at the centre of curvature of the mirror. Therefore, in this case, the focus of the lens and the centre of curvature of the mirror will coincide.
If you look at the figure, the radius of curvature R is equal to the sum of focal length of the lens and the distance of separation.
Hence, $R=|{{f}_{1}}|+d$ … (1).
And we know that for a concave mirror $R=2|f|$
Hence, $R=2|{{f}_{2}}|$
Substitute the value of R in equation (1)
$\Rightarrow 2|{{f}_{2}}|=|{{f}_{1}}|+d$
$\Rightarrow d=2|{{f}_{2}}|-|{{f}_{1}}|$