Question

The size of the image of an object, at infinity, is formed by a convex lens of focal length $30\,cm$ is $2\,cm$. If the concave lens of focal length $20\,cm$ is placed in between the convex lens and the image, at a distance of $26\,cm$ from the convex lens, calculate the new size of the image.
A. 1.25 cm
B. 2.5 cm
C. 1.05 cm
D. 2 cm

Answer 1

Hint:Now in this question the second lens (concave lens) is placed in between the lens and the image formed by the lens. So, in such scenarios the image will act as the object for the second lens and the second lens will form the image accordingly. This way of solving such problems is always valid in any condition.

Formula used:
Lens formula,
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
where $v$ is the distance of the image, $u$ is the distance of the object and $f$ is the focal distance of the lens.
Magnification formula,
\[m = \dfrac{v}{u}\]
Where $v$ is the distance of the image, $u$ is the distance of the object and $m$ is the magnification.

Complete step by step answer:
Let the focal length of the first lens (convex lens) $ = {f_1} = 30\,cm$. Now the object is at infinity so the image is formed at the focal length of the convex lens. Which is 30 cm from the lens. Let the focal length of the second lens (concave lens) $ = {f_2} = - 20\,cm$ as a concave lens has negative focal length.

Now the distance between the concave lens and the image formed $ = 30\,cm - 26\,cm = 4\,cm$ away from the lens. Now we have,
${u_2} = 4\,cm$ and ${f_2} = - 20\,cm$
And we need to find the distance of the image formed by the second lens.
$\dfrac{1}{{{v_2}}} - \dfrac{1}{{{u_2}}} = \dfrac{1}{{{f_2}}}$

Let’s substitute the value.
$\dfrac{1}{{{v_2}}} - \dfrac{1}{4} = \dfrac{1}{{ - 20}}$
$\Rightarrow \dfrac{1}{{{v_2}}} = \dfrac{1}{4} - \dfrac{1}{{20}}$
$\Rightarrow \dfrac{1}{{{v_2}}} = \dfrac{{5 - 1}}{{20}}$
We took the L.C.M of the denominators which is 20
$\dfrac{1}{{{v_2}}} = \dfrac{4}{{20}} \\
\Rightarrow \dfrac{1}{{{v_2}}} = \dfrac{1}{5}$
$\Rightarrow {v_2} = 5\,cm$

Now we need to find the new size (magnification) of the image.
So, in this case magnification is,
\[m = \dfrac{{{v_2}}}{{{u_2}}} \\
\Rightarrow m = \dfrac{5}{4} \\
\Rightarrow m = 1.25\]
The actual height of the image formed by the convex lens $ = 2cm$
Now the height of the new image formed by the concave lens $ = m \times {h_o}$ where $m$ is the magnification and ${h_o}$ is the height of the object.
So, the height of the new image formed by the concave lens $ = m \times {h_o} = 1.25 \times 2 = 2.5cm$
The height of the final image $ = 2.5\,cm$

Therefore, the correct option is B.

Note:In this case we assume that the light rays start from the negative side of the lens and move towards the positive side of the lens meaning the starting side of light marks the side as the negative side. Thus, the object or image in the negative side of the lens (the side of the lens where light rays started to move towards the lens) will have a negative distance and the object or image in the positive side of the lens (the remaining other side of the lens) will have a positive distance.