Find the longitudinal magnification of an object of finite width which is placed at $20\,cm$ left of an diverging lens of focal length $ - 30\,cm$.
A) $0.25$
B) $0.30$
C) $0.36$
D) $0.4$

Hint: In optics, magnification refers to an increase in the size of the image in proportion to the size of the object that created the image. Diverging lenses direct light away from the optical centre or lens axis. Concave lenses always work like diverging lenses.

Formula used:
\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]
Complete step by step answer:
First of all, we apply the lens formula to find the distance of the image \[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]
\[ \Rightarrow \dfrac{1}{v} = \dfrac{1}{u} + \dfrac{1}{f}\]
It is given that the question stated as, $u = - 20$ and $f = - 30$
Now putting the values of u and f get in the lens formula we get
\[\dfrac{1}{v} = \dfrac{1}{{ - 30}} + \dfrac{1}{{ - 20}}\]
$ \Rightarrow \dfrac{1}{v} = \dfrac{{ - 2 - 3}}{{60}}$
$ \Rightarrow \dfrac{1}{v} = - \dfrac{5}{{60}}$
So finally our position of objects comes
$v = - 12$
We know that there are two types of magnification transverse and lateral.

Transverse magnification:
Linear magnification (also known as Transverse magnification) refers to the ratio of the length of an image to the length of an object, measured in a plane perpendicular to the optical axis.
${{\text{m}}_{\text{T}}}{\text{ = }}\dfrac{{{\text{Height}}\,{\text{of image}}\left( {{{\text{h}}_{\text{I}}}} \right)}}{{{\text{Height}}\,{\text{of}}\,{\text{object}}\left( {{{\text{h}}_{\text{o}}}} \right)}}$
${m_T} = \dfrac{{{h_I}}}{{{h_o}}} = \dfrac{v}{u} = \dfrac{f}{{f + u}}$

Longitudinal magnification:
Longitudinal magnification refers to the factor due to which the measured image along the optical axis becomes larger.
${{\text{m}}_{\text{L}}}{\text{ = }}\dfrac{{{\text{Length}}\,{\text{of}}\,{\text{image}}\left( {{{\text{l}}_{\text{I}}}} \right)\,}}{{{\text{Length}}\,{\text{of}}\,{\text{object}}\left( {{{\text{l}}_{\text{o}}}} \right)}}$
$m{}_{_L} = \dfrac{{{l_I}}}{{{l_O}}} = \dfrac{{\left| {{v_2}} \right| - \left| {{v_1}} \right|}}{{\left| {{u_2}} \right| - \left| {{u_1}} \right|}}$
For small object the relationship between the two magnifications is
${m_L} = \dfrac{{{v^2}}}{{{u^2}}} = {m_T}^2$
Longitudinal Magnification $ = \dfrac{{{v^2}}}{{{u^2}}}$
$ \Rightarrow {m_L} = {\left( {\dfrac{v}{u}} \right)^2} = {\left( {\dfrac{{12}}{{20}}} \right)^2}$
$ \Rightarrow {m_L} = {\left( {\dfrac{6}{{10}}} \right)^2}$
${m_L} = 0.36\,cm$

Hence the correct option is (C).

Note: While solving the problem always keeps in mind that distance in the direction of a light ray is taken positive while in the opposite direction distance is negative. A diverging lens, also known by the name of a negative lens. If the refracted beam goes beyond the point, then the lens is off. This is observed when creating a virtual image. It can be used in cases where the light intensity needs to be delivered to the desired area. There is one more type of magnification known as angular magnification. Angular magnification is the ratio of a tangent to an angle defined by an object and an image measured at a single point on the device.