Question

An object is kept at a distance of 16 cm from a thin lens and the image formed is real. If the object is kept at a distance of 6 cm from the same lens, then the image formed is virtual. If the sizes of the image formed are equal, then the focal length of the lens will be
(A) 15 cm
(B) 17 cm
(C) 21 cm
(D) 11 cm

Answer 1

Hint: The given lens is a convex lens therefore the object distance (u) will be negative. We can use the definition of magnification here to calculate the value of the focal length. From the formula \[\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\] and the formula for magnification we need to find the focal length by substituting the value of the magnification and the object distance from the question we can find the focal length.
Formula used:
In the solution we will be using the following formula,
\[\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\]
Where \[f\] is the focal length
\[v\] is the distance from the image
\[u\] is the distance from the object

Complete step by step solution:
It must be a convex lens as both the real and virtual image are formed by lens.
We have,
\[\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\]
By multiplying the object distance on both the sides we get,
\[\dfrac{u}{v} + 1 = \dfrac{u}{f}\]
We know that \[v = - mu\]
Therefore the above equation becomes,
\[ \Rightarrow - \dfrac{1}{m} + 1 = \dfrac{u}{f}\]
\[ \Rightarrow m = \dfrac{f}{{f + u}}\]….. (1)
Now, let's consider first that image formed is real
Therefore, \[u = - 16\] cm and magnification \[m = - 1\]
Putting the values of \[m\] and \[u\] in equation (1)
\[ \Rightarrow - 1 = \dfrac{f}{{f + ( - 16)}}\]…. (2)
Now, let us consider first that image formed is virtual
Therefore, \[u = - 6\] and magnification \[m = 1\]
Putting the values of \[m\] and \[u\] in equation (1)
\[ \Rightarrow 1 = \dfrac{f}{{f + ( - 6)}}\]….. (3)
Therefore from equation (2) and (3) we get,
\[ - \dfrac{f}{{f - 16}} = \dfrac{f}{{f - 6}}\]
\[ \Rightarrow - (f - 6) = f - 16\]
Therefore on further calculation we get,
\[ \Rightarrow \]\[2f = 22\]
\[ \Rightarrow \]\[f = 11cm\]
Therefore, if the size of the image formed are equal, then the focal length of the lens will be 11 cm

Therefore, option (D) is the correct answer.

Note: The magnification will be negative if the image is real and positive if the image is virtual. We should always be careful with the sign convention as we can make a mistake here.