Question

An extended object of size $2\;{\text{mm}}$ is placed on the principal axis of a converging lens of focal length $10\,{\text{cm}}$. It is found that when the object is placed perpendicular to the principal axis the image formed is $4\;{\text{mm}}$ in size. The size of image when it is placed along the principal axis is____________mm

Answer 1

Hint:The ratio of size of the image to the size of the object. First the magnification factor of the lens has to be found by the case of when the condition of the object is placed in the perpendicular to the principal axis. By using that calculated value, the size of the image placed along the principle axis can be calculated.

Useful formula:
When object is placed perpendicular to the principal axis then,
${I_{size}} = m \times {O_{size}}$
Where ${I_{size}}$is the size of the image, $m$is the magnification factor and ${O_{size}}$is the size of the object

When the object is placed on the principle axis then,
${\text{Size}}\;{\text{of}}\;{\text{the}}\;{\text{image}}\;\left( {{{\text{I}}_{size}}} \right) = {m^2} \times object\;size\left( {{{\text{O}}_{size}}} \right)$
where $m$is the magnification factor.

Complete step by step answer:
Given, Size of the object is ${O_{size}} = 2\;{\text{mm}}$
Size of the image ${I_{size}} = 4\;{\text{mm}}$

When object is placed perpendicular to the principal axis then,
${I_{size}} = m \times {O_{size}}$
Substitute all the values in the above equation.
$
  4\;mm = m \times 2\,mm \\
  m = 2 \\
$
When the object is placed on the principle axis then.
${\text{Size}}\;{\text{of}}\;{\text{the}}\;{\text{image}}\;\left( {{{\text{I}}_{size}}} \right) = {m^2} \times object\;size\left( {{{\text{O}}_{size}}} \right)$
Substitute all the values in the above equation.
$
  {\text{Size}}\;{\text{of}}\;{\text{the}}\;{\text{image}}\;\left( {{{\text{I}}_{size}}} \right) = {2^2} \times 2\;{\text{mm}} \\
  {{\text{I}}_{size}} = 4 \times 2\;{\text{mm}} \\
  {{\text{I}}_{size}} = 8\;{\text{mm}} \\
 $

Thus, the size of the image when object principle axis is ${I_{size}} = 8\;{\text{mm}}$


Note:
When an object is placed perpendicular to the principal axis, then the magnification factor is the ratio of the size of the image to the object's size. However, for the case of when an object is placed along the principal axis, then the magnification factor is the ratio of square root value of size of the image and to the size of the object.