Question

An object is placed $12cm$ to the left of a converging lens of focal length $8cm$.Another converging lens of $6cm$ focal length is placed at a distance of $28cm$ to the right of the first lens. Therefore the second lens will produce image which will be,
A. real, enlarged
B. virtual, enlarged
C. real, diminished
D. virtual, diminished

Answer 1

The lens formula is the basic formula in order to find the answer for this question. That is, the reciprocal of the focal length will be equal to the reciprocal of the image formed distance minus the reciprocal of the object distance. These all may help us to solve this question.

Complete step by step answer:
First of all let us look at the first lens.
Here the initial object distance is given as,
${{u}_{1}}=-12cm$
The focal length of the first lens is mentioned as,
${{f}_{1}}=8cm$
For the second lens, the focal length is given as,
${{f}_{2}}=6cm$
The distance between the first and the second lens is given as,
$d=28cm$
First of all let us use the lens formula for the first lens.
$\dfrac{1}{{{f}_{1}}}=\dfrac{1}{{{v}_{1}}}-\dfrac{1}{{{u}_{1}}}$
Rearranging the equation in terms of velocity will give,
$\dfrac{1}{{{v}_{1}}}=\dfrac{1}{{{f}_{1}}}+\dfrac{1}{{{u}_{1}}}$
Substituting the values in it,
$\dfrac{1}{{{v}_{1}}}=\dfrac{1}{8}+\dfrac{1}{-12}=\dfrac{4}{96}=\dfrac{1}{24}$
Therefore the image formed by the first lens is found to be a virtual image which is enlarged in nature. Therefore the correct answer is given as option
${{v}_{1}}=24cm$
Hence we can see that the image is formed at the right side at a distance of $24cm$. This image will be acting as the object for the second lens. The object distance for the second lens will be the difference of the distance from the lens and the distance of image formed by the first lens.
Therefore we can write that,
The object distance of the second lens is,
${{u}_{2}}=28-24cm=-4cm$
The focal length of second lens is,
${{f}_{2}}=6cm$
Therefore the object is placed in between the focus of the lens and the lens.

So, the correct answer is “Option B”.

Note: According to the sign convention, the direction of all the distances which is parallel to the incident light rays will be taken as positive. And all the distances antiparallel to the incident light rays are taken as negative. This is to be followed strictly when we solve the question.