Question

A beam of parallel rays is brought to a focus by a convex lens. If a thin concave lens of equal focal length is joined to the convex lens, the focus will
A. Be shifted to infinity.
B. Be shifted by a small distance
C. Remain undisturbed
D. None of the above

Answer 1

Use the formula for focal length of combination of two lenses. We know that focal length of convex lens is positive and focal length of concave lens is negative.

Formula used:
The focal length of combination of two lenses is,
$\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
Here, ${f_1}$ and ${f_2}$are the focal lengths of two lenses.

Complete step by step answer:
We know that when two lenses of focal length ${f_1}$ and ${f_2}$ are joined together, the focal length of the combination is can be calculated by the formula,
$\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
Here, f is the focal length of combined lenses.
We know that the focal length of a convex lens is positive while the focal length of a convex lens is negative.
We have given that the focal length of both convex and concave lenses is equal. Therefore, we can write the above equation as,
$\dfrac{1}{f} = \dfrac{1}{f} + \dfrac{1}{{\left( { - f} \right)}}$
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{f} - \dfrac{1}{f}$
$\Rightarrow \dfrac{1}{f} = 0$
$\Rightarrow f = \dfrac{1}{0}$
$\therefore f = \infty$
Therefore, the focus of the combination of convex and concave lens becomes infinite.
To remain the focus undisturbed, the focal length of the concave lens must be zero. But the focal length of a concave lens can never be zero. Therefore, the option (C) is incorrect.

So, the correct answer is “Option A”.

Note:
To solve the questions relating to the combination of lenses, students should remember the sign of focal lengths of convex lens and concave lens. The focal length of a convex lens is positive while the focal length of a convex lens is negative. Note that, convex lenses converge the parallel rays falling on it while concave lenses diverge the parallel rays falling on it.