Question

Image of an object at infinity is formed by a convex lens of focal length 30 cm such that the size of image is 2cm. If the concave lens of focal length 20cm is placed in between the convex lens and the image, at a distance of 26cm from the convex lens, the size of new Image is-
A. 2.5cm
B. 2.0cm
C. 1.025cm
D. 1.05cm

Answer 1

Hint: When a lens is already making an image and another lens (whatever the type) is placed in between the image and the previously placed lens then the image formed by the previously placed lens will act as an object for the lens which is placed in between.

Complete step by step answer:
Given,
Focal length of convex lens=\[{f_1}\]=30cm
And is it also given that the object is at infinity so the image formed by the convex lens is at focal point.
So the image distance for the convex lens is the same as that of the focal length of the convex lens.
So the image distance for the convex lens =${v_1}$=${f_1}$=30cm
And the object distance for the convex lens= Infinity
Now the concave lens is placed in between the convex lens and the image formed by that, so this image will now be acting like an object for the concave lens.
Since the concave lens is placed at 26 cm from the convex lens.
So for the concave lens object distance $ = {u_2}$= 30-26=4cm
And the focal length of the concave lens = ${f_2}$=-20cm
So for calculation of the image distance for the concave lens,
By using the lens formula for the concave lens,
$\dfrac{1}{{{v_2}}} - \dfrac{1}{{{u_2}}} = \dfrac{1}{{{f_2}}}$
Now substituting the values from the given,
 $ \Rightarrow \dfrac{1}{{{v_2}}} - \dfrac{1}{4} = \dfrac{1}{{ - 20}}$
${v_2} = 5cm$
Now we will find the magnification for the concave lens as follows,
$m = \dfrac{{{v_2}}}{{{u_2}}}$
Substituting the values we get
$m = \dfrac{5}{4}$--equation (1)
Also there is another formula for the magnification
$m = \dfrac{{{h_i}}}{{{h_0}}}$--equation (2)
We have given that the height of the image formed by the convex lens is 2cm,this is an object for the concave lens so for the concave lens the object size is,
${h_0} = 2cm$, now we have to calculate the image size for the concave lens ${h_i} = ?$.
 Now equating equation (1) and equation (2) and substituting the values, we get the following,
$\dfrac{5}{4} = \dfrac{{{h_i}}}{2}$
$\
\Rightarrow {h_i} = \dfrac{5}{4} \times 2 \\
\Rightarrow {h_i} = \dfrac{5}{2} \\
\Rightarrow {h_i} = 2.5cm \\$
Hence the size of new image$ = 2.5cm$

So the correct option is (A).


Note:
In a convex lens, if the object is placed at focal point then its image will be at infinity and the image of the object placed at infinity will be at focal point. This is a quite interesting and useful concept used in optics.