Question

A convex lens produces an inverted image magnified three times of an object placed at a distance of $15cm$ from it. Calculate focal length of the lens.

Answer 1

Hint: The magnification is the ratio of height of the image to the height of the object. In the problem it is mentioned that magnification is three times, that is height of the image is three times the height of the object.

Formulas used:
Lens formula:
\[\dfrac{1}{f} = \dfrac{1}{V} - \dfrac{1}{U}\] …………. $\left( 1 \right)$
Magnification:
$m = \dfrac{V}{U}$ ……….. $\left( 2 \right)$
Where, $U = $ Object distance.
$V = $ Image distance.
$f = $ Focal length
$m = $Magnification.

Complete step-by-step solution:
Given:-
Magnification, $m = - 3$ (Negative sign because the image is inverted).
Object distance, $U = - 15cm$ (Negative sign because the object is kept in negative x-axis).
Focal length, $f = ?$
We know that, magnification of the mirror is given by
$m = \dfrac{V}{U}$
Substituting the given data in above equation, we get
$ - 3 = \dfrac{V}{{\left( { - 15} \right)}}$
On simplifying the above equation,
$V = 45cm$ (Positive sign because the image obtained be on a positive x-axis).
And also, lens formula is given by
\[\dfrac{1}{f} = \dfrac{1}{V} - \dfrac{1}{U}\]
On substitution,
 \[\Rightarrow \dfrac{1}{f} = \dfrac{1}{{45}} - \dfrac{1}{{ - 15}}\]
\[\Rightarrow \dfrac{1}{f} = \dfrac{1}{{45}} + \dfrac{1}{{15}}\]
\[\Rightarrow \dfrac{1}{f} = \dfrac{{1 + 3}}{{45}}\]
\[\Rightarrow \dfrac{1}{f} = \dfrac{4}{{45}}\]
Therefore, focal length $f = 11.25cm$.
Additional information:
 Sign convention:
$ \Rightarrow $ Object distance = Negative
$ \Rightarrow $ Image distance = Positive if the image is real and negative if the image is virtual
$ \Rightarrow $ Focal length = Positive
$ \Rightarrow $ Height of object = Positive
$ \Rightarrow $ Height of image = Negative if the image is real and positive if the image is virtual
$ \Rightarrow $ Magnification = Negative if the image is real and positive if the image is virtual.

Note: It should be noted that convex lenses can produce images which can be either real or virtual. When the nature of the image is real the image obtained will be inverted and if the nature of the image is virtual the image obtained will be erect and the proper sign convention should be followed to solve the problem.