Question

In the displacement method, a convex lens is placed in between an object and a screen. If one of the magnitude is 3 and the displacement of the lens between the two positions is 24cm, then the focal length of the lens is:
$
  (a){\text{ 10cm}} \\
  (b){\text{ 9cm}} \\
  (c){\text{ 6cm}} \\
  (d){\text{ 16/3cm}} \\
 $

Answer 1

Hint: In this question let the distance of the image from lens be v cm and the object distance be u cm. Use the direct formula for the magnification of the lens that is $M = \dfrac{v}{u}$. Then use the constraints of the questions and the fact that the difference of the distance of the image from the lens and distance of the object from the lens is given. This will help approach the solution.

Complete step-by-step solution -
Let the distance of the image from the lens be v cm.
And the distance of the object from the lens be u cm.
Now as we all know that the magnification of the lens is the ratio of the distance of the image from the lens to distance of the object from the lens, often denoted by M.
$ \Rightarrow M = \dfrac{v}{u}$
And it is given that the magnification is three.
$ \Rightarrow M = \dfrac{v}{u} = 3$
$ \Rightarrow v = 3u$....................... (1)
Now it is also given that the displacement of the lens between the two positions is 24cm.
I.e. it is the difference of the distance of the image from the lens and distance of the object from the lens.
$ \Rightarrow v - u = 24$
Now substitute the value of v from equation (1) in the above equation we have,
$ \Rightarrow 3u - u = 24$
Now simplify this we have,
$ \Rightarrow 2u = 24$
$ \Rightarrow u = \dfrac{{24}}{2} = 12$cm
Now from equation (1) we have,
$ \Rightarrow v = 3u = 3\left( {12} \right) = 36$cm.
Now according to the lens formula we have,
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Where, f = focal distance of the lens, v = image distance and u = object distance.
Now substitute the value we have,
$ \Rightarrow \dfrac{1}{f} = \dfrac{1}{{36}} + \dfrac{1}{{12}}$
Now simplify this by taking L.C.M we have
$ \Rightarrow \dfrac{1}{f} = \dfrac{{1 + 3}}{{36}} = \dfrac{4}{{36}} = \dfrac{1}{9}$
So the focal length of the lens is 9 cm.
So this is the required answer.
Hence option (b) is the correct answer.

Note: Magnification as the name suggests refers to the process of magnifying or enlarging in order to have a more clearer visual for a distance object. The image formed can be magnified too in some cases. The trick point here was that since we are dealing with a convex lens, the focal length of the convex lens is always positive.