Question

A diverging lens of focal length 20cm forms an image 15cm from the lens. What is the distance of the object from the lens?
(A) 20 cm
(B) 15 cm
(C) 60 cm
(D) 30 cm

Answer 1

Consider the image formed to be at 15 cm from either side of the lens, and then solve the lens equation to find the distance of the object from the lens.Lenses are basically magnifying glasses with curved sides. A lens is a piece of transparent glass which concentrates or disperses light rays when it passes through them by refraction. Due to the magnifying property, lenses are used in telescopes and other magnifying devices. They are employed in cameras for gathering the light rays.

Complete step by step answer:
There is a diverging lens of focal length 20cm which forms an image 15cm from the lens. We must find out the distance of the object from the lens whose images formed. An image is formed through a lens by refraction from it. A lens has two refracting surfaces, and both are curved surfaces. There are two types of lenses: a diverging lens and a converging lens. A converging lens converges the light rays parallel to the principal axis of the lens, to the point called focus of the lens, and distance of the lens from the focus is called focal length. A diverging lens diverges the light rays parallel to the principal axis of the lens in such a way that it looks like it comes out from an imaginary object at the focus of the lens. The distance of the object from the lens (u), the distance of image from the lens (v) and the distance of focus from the lens (f) are related to each other by the lens equation
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

A general sign convention is followed while doing calculations i.e. the right-hand side of the lens is positive and the left-hand side of the lens is the negative side. The distances measured follow this sign convention according to their position from the lens. The focal length of a diverging lens is taken as negative and that of a converging lens is taken as positive. Since in the question it is not mentioned about the side of the lens where the image is formed, we can consider both the sides and then check which option matches the answer. If the image is on the left side then u = -15cm, otherwise u = 15cm,f = -20cm
Right side:
\dfrac{1}{v} - \dfrac{1}{{15}} = \dfrac{1}{{ - 20}} \\\ \Rightarrow\dfrac{1}{v} = \dfrac{1}{{15}} - \dfrac{1}{{20}} = \dfrac{{4 - 3}}{{60}} = \dfrac{1}{{60}} \\\ \therefore v = 60cm \\\
Left side:
\dfrac{1}{v} - \dfrac{1}{{ - 15}} = \dfrac{1}{{ - 20}} \\\ \Rightarrow\dfrac{1}{v} = - \dfrac{1}{{15}} - \dfrac{1}{{20}} = \dfrac{{ - 4 - 3}}{{60}} = \dfrac{{ - 7}}{{60}} \\\ \therefore v = - 8.57cm \\\

Hence,option (C) is the correct answer.

Note: While solving the problems of lenses and geometrical optics, always keep in mind to consider the proper sign conventions otherwise you might end up getting a wrong answer.All the distances that are measured along the direction of incident rays are taken as positive and vice-versa.