Question

A convex lens makes a real image $4\,\,cm$ long on a screen. When the lens is shifted to a new position without disturbing the object or the screen, we again get a real image on the screen which is $9\,\,cm$ long. The length of the object must be:
(A) $2.5\,\,cm$
(B) $6\,\,cm$
(C) $6.5\,\,cm$
(D) $36\,\,cm$

Answer 1

Hint: The above problem can be solved using the formula of the magnification of the lenses in the correlation with the formula of the height of the object. A convex lens is also referred to as the converging lens, because it is a lens that converges rays of light that are advancing in parallel to its principal axis.

Formulae Used:
The magnification of the lens:
$m = \dfrac{v}{u}$
Where, $m$ denotes the magnification, $v$ denotes the distance of the image, $u$ denotes the distance of the object.

Complete step-by-step solution:
The data given in the problem are:
Image length that forms on the screen $4\,\,cm$,
After moving the lens image length that forms on the screen $9\,\,cm$.
Let the height of the object be ${h_1}$.
On the first position of the lens:
Magnification the convex lens:
 ${m_1} = \dfrac{4}{{{h_1}}} = - \dfrac{v}{u}$
At the second position of the lens:
Magnification the convex lens:
${m_2} = \dfrac{9}{{{h_1}}} = - \dfrac{u}{v}$
Now $m = {m_1} \times {m_2}$
$\left( { - \dfrac{v}{u}} \right) \times \left( { - \dfrac{u}{v}} \right) = \left( {\dfrac{4}{{{h_1}}}} \right) \times \left( {\dfrac{9}{{{h_1}}}} \right)$
Equating the above equation:
$h_1^2 = 36$
Taking square root on both sides:
 ${h_1} = 6\,\,cm$
Therefore, the length of the object is ${h_1} = 6\,\,cm$.
Hence the option (B) ${h_1} = 6\,\,cm$ is the correct answer.

Note:- The linear magnification is also in relation to the distance of the object and distance of the object. This shows that linear magnification created by a lens is also identical to the ratio of the image distance to the object distance. Magnification has no unit. The reason behind it is that it is a ratio of the measurement that It is the ratio of the size of an image to the size of an object.