Question

A $2cm$ tall object is placed perpendicular to the principal axis of a convex lens of focal length $10cm$. The distance of the object from the lens is $15cm$ find the nature, position and size of the image and also find its magnification.

Answer 1

Hint We are given with the height of the object, the object distance and the focal length of the lens and are asked to find the image distance, magnification and the height of the image. Thus, we will use the formulas for finding these values taking into consideration the sign convention for lenses. We will use the lens formula and the formula for linear magnification for a lens.
Formulae Used:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Where, $f$ is the focal length of the lens, $v$ is the image distance and $u$ is the object distance.
$m = \dfrac{{{h_i}}}{{{h_o}}} = \dfrac{v}{u}$
Where, $m$ is the linear magnification by the lens, ${h_i}$ is the height of the image and ${h_o}$ is the height of the object.

Complete Step By Step Solution
Here,
The lens is a convex one.
Thus, focal length is positive
Thus,
$f = + 10cm$
The object is placed in front of the lens
Thus,
$u = - 15cm$
Now,
Applying the lens formula,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Further, we get
$\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$
Thus, we get
$\dfrac{1}{v} = \dfrac{1}{{ + 10}} + \dfrac{1}{{( - 15)}}$
Further, we get
$\dfrac{1}{v} = \dfrac{1}{{10}} - \dfrac{1}{{15}}$
Taking LCM and calculating, we get
$\dfrac{1}{v} = \dfrac{{3 - 2}}{{30}}$
Thus, we get
$v = + 30cm$
Now,
Applying the formula for linear magnification
$m = \dfrac{v}{u} = \dfrac{{ + 30}}{{ - 15}}$
Further, we get
$m = - 2$
Again,
$m = \dfrac{{{h_i}}}{{{h_o}}}$
Now,
The object is upright above the principal axis.
Thus,
${h_o} = + 2cm$
Thus,
$- 2 = \dfrac{{{h_i}}}{{ + 2}} \Rightarrow {h_i} = - 4cm$
Hence, the image is formed $30cm$ from the lens. The image formed is real and inverted in nature. The size of the image is $4cm$ and the lens produces a linear magnification of $- 2$.

Note The value of magnification is negative and thus the image formed is real. The value of the magnification is greater than one thus the image will be enlarged. The value of height is negative which signifies that the image is inverted that means it is below the principal axis.