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Question: The focal length of a convex lens is \[{\text{2}}{\text{.5 cm}}\]. Its magnifying power for minimum ...

The focal length of a convex lens is 2.5 cm{\text{2}}{\text{.5 cm}}. Its magnifying power for minimum distance of distinct vision will be:
A. 2525
B. 5252
C. 1111
D. 1.11.1

Explanation

Solution

First of all, we will find the expression for the lens, which involves image distance, object distance and the focal length. The object distance is the least distance of distinct vision. We will substitute the required values in the expression and manipulate accordingly to obtain the result.

Formula used:
The relation between u, v and f for a convex lens is:
1f=1u1v\dfrac{1}{f} = \dfrac{1}{u} - \dfrac{1}{v}
Where, ff is Focal length of a convex lens, vv is the distance of the image from the lens, uu is the distance of the object from the lens.

Complete step by step solution:
Given: Focal length of a convex lens (f) = 2.5 cm2.5{\text{ }}cm
vv is the distance of the image from the lens.
u=  25 cmu = {\text{ }} - {\text{ }}25{\text{ }}cm
the relation between u, v and f for a convex lens is:
1f=1u1v\dfrac{1}{f} = \dfrac{1}{u} - \dfrac{1}{v} …………(i)
Where f is Focal length of a convex lens
vv = the distance of the image from the lens
uu = the distance of the object from the lens
Equation (i):
1/ 2.5 = 1/25  1/u1/{\text{ }}2.5{\text{ }} = {\text{ }}1/25{\text{ }} - {\text{ }}1/u
1/u = 1/25  1/ 2.5\Rightarrow 1/u{\text{ }} = {\text{ }} - 1/25{\text{ }} - {\text{ }}1/{\text{ }}2.5
1/u = 1/25  10/25\Rightarrow 1/u{\text{ }} = {\text{ }} - 1/25{\text{ }} - {\text{ }}10/25
1/u = 11/25\Rightarrow 1/u{\text{ }} = {\text{ }} - 11/25
u = 25/11\Rightarrow u{\text{ }} = {\text{ }} - 25/11
Hence the distance of the object from the lens is 25/11 - 25/11.The magnifying power, or extent to which the object being viewed appears enlarged, and the field of view, or size of the object that can be viewed, are related by the geometry of the optical system.

Magnifying power (M) = v/u  Magnifying{\text{ }}power{\text{ }}\left( M \right){\text{ }} = {\text{ }} - v/u\;
Magnifying power (M)  = 25×1125Magnifying{\text{ }}power{\text{ }}\left( M \right)\; = {\text{ }}\dfrac{{ - 25 \times 11}}{{ - 25}}
Magnifying power (M) = 11
Hence magnifying power for minimum distance of distinct vision will be: 11

Hence, option C is the correct answer.

Note: Magnification is a different term from magnifying power. Magnification is equal to the ratio of size of image and size of object. Whereas magnifying power is equal to the ratio of the dimension of the image and the object. While solving the problem, most of the students make mistakes in choosing signs for the focal length. Since, the convex lens is a converging lens, its focal length is positive and negative in case of a concave lens.