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Question: A convex lens of \(2D\) power is joined with a concave mirror of \(1D\) power. Equivalent power of i...

A convex lens of 2D2D power is joined with a concave mirror of 1D1D power. Equivalent power of instrument will be
A. −3D - 3D
B. +3D + 3D
C. −5D - 5D
D. +5D + 5D

Explanation

Solution

Hint:- Though it is known that lenses and mirrors are different, but convex lenses and concave mirrors show some similarities sometimes. One of the properties of the convex lens and concave mirror is that both of them converge parallel rays to the focal point and diverge parallel rays away from the focal point.

Complete step-by-step solution
Step I:
For lens mirror combination, the relation between the focal length of the lens and mirror is given as
1fequivalent=2flens−1fmirror\dfrac{1}{{{f_{equivalent}}}} = \dfrac{2}{{{f_{lens}}}} - \dfrac{1}{{{f_{mirror}}}}
Step II:
Also focal length is defined as the inverse of the power of the instrument. Therefore focal length of the optical instrument is written as
f=1Powerf = \dfrac{1}{{Power}}
Where f is the focal length
Given the power of lens is =2D = 2D
Also the power of mirror is=1D = 1D
Substitute value of the power of lens in the above formula,
flens=12{f_{lens}} = \dfrac{1}{2}
1flens=2m\dfrac{1}{{{f_{lens}}}} = 2m
Step III:
Power of a concave mirror is always positive, therefore
fmirror=11{f_{mirror}} = \dfrac{1}{1}
1fmirror=1m\dfrac{1}{{{f_{mirror}}}} = 1m
Step III:
Substitute the value in the formula,
1fequivalent=2(2)−1\dfrac{1}{{{f_{equivalent}}}} = 2(2) - 1
=4−1= 4 - 1
1fequivalent=3\dfrac{1}{{{f_{equivalent}}}} = 3
fequivalent=13{f_{equivalent}} = \dfrac{1}{3}
Step IV:
It can be written that
P=1fequivalentP = \dfrac{1}{{{f_{equivalent}}}}
P=113P = \dfrac{1}{{\dfrac{1}{3}}}
Power of an optical instrument is always measured in dioptres. It is denoted by ‘D’.
P=3DP = 3D
Step V:
Hence, the equivalent power of the instrument will be 3D3D.
Therefore, Option B is the right answer.

Note:- Since power and focal length are inversely related therefore, it is to be remembered that an optical instrument with shorter focal length will be more powerful. Power of an optical instrument is the measure of the convergence or divergence produced by the instrument. It is to be noted that the lens when dipped in water will have a focal length of infinity. Thus the focal length will always be zero.