Question

A convex lens of 2 D power is joined with a concave mirror of 1 D power. The equivalent power of instrument will be
A. + 3 D
B. – 5 D
C. – 3 D

Answer 1

Hint: The power of lens and a mirror is equal to the reciprocal of the focal length of the lens and the mirror respectively. By first obtaining the equivalent focal length of the combination of mirror and lens, we can convert its power which will be the required answer.

Formula used:
The power of a lens or a mirror is related to the focal length of lens or a mirror by the following relation:
$P = \dfrac{1}{f}$
The equivalent focal length of a combination of a convex lens and a concave mirror is given as
$\dfrac{1}{{{f_{equivalent}}}} = \dfrac{1}{{{f_{mirror}}}} - \dfrac{2}{{{f_{lens}}}}$

Complete step by step answer:
We are given a convex lens whose power is given as
${P_{lens}} = 2D$
Therefore, the focal length of the convex lens is
${f_{lens}} = \dfrac{1}{{{P_{lens}}}} = \dfrac{1}{2}m$
This lens is joined with a concave lens whose power is given as
${P_{mirror}} = 1D$
Therefore, its focal length is given as
$\Rightarrow {f_{mirror}} = \dfrac{1}{{{P_{mirror}}}} = 1m$
Now when a convex lens is joined with a concave mirror, we obtain a convex mirror. The equivalent focal length of this combination is given as
$
\Rightarrow \dfrac{1}{{{f_{equivalent}}}} = \dfrac{1}{{{f_{mirror}}}} - \dfrac{2}{{{f_{lens}}}} \\
\Rightarrow \dfrac{1}{{{f_{equivalent}}}} = 1 - 2 \times 2 = 1 - 4 = - 3 \\
$
Now, we can get the power of this combination which can be obtained by taking the reciprocal of the focal length and is given as
$\Rightarrow {P_{equivalent}} = \dfrac{1}{{{f_{equivalent}}}} = - 3D$
Therefore the power of the obtained convex mirror is – 3 D. Hence, the correct answer is option C.

Note:
1. All the distances to the left of a mirror and a lens are taken as negative while those to the right of the lens and the mirror are taken as positive.
2. The convex mirror has negative focal length due to which its power is also negative.
3. The formula for obtaining the equivalent focal length can be converted into formula for powers by simply substituting the reciprocal of the focal length as the power. Doing so, we get
${P_{equivalent}} = {P_{mirror}} - 2{P_{lens}}$