Question

A converging mirror forms a real image of height $4cm$ of an object of height $1cm$ placed $20cm$ away from the mirror. Calculate focal length of the mirror.

Answer 1

Hint:Use the formula for magnification in terms of image and object heights. This will give you the value of \[v\]. Use this value to find the image distance, that is, \[v\]. As the focal length is given, use the mirror formula to get the value of \[f\]. Use suitable sign conventions.

Complete step by step answer:
 It is given that the height of the image ${h_2}$ = 4 cm
Height of the object ${h_1}$ = 1 cm
Magnification of a mirror is defined as the ratio of the height of an image to the height of an object. Therefore,
\[m = \dfrac{{\left| {{h_2}} \right|}}{{\left| {{h_1}} \right|}}\]
\[ \Rightarrow m = \dfrac{{ - 4}}{1} \\
\Rightarrow m = - 4\]
Also, Magnification of the mirror is equal to the ratio of image distance to that of object distance. Therefore , \[m = \dfrac{{\left| v \right|}}{{\left| u \right|}}\] where, u is the object distance and v is the image distance.
\[ m = \dfrac{{ - v}}{{ - 20}} \\
\Rightarrow m = \dfrac{v}{{20}}\]

Equating both the equations we get, \[ - 4 = \dfrac{v}{{20}} \Rightarrow v = - 80\,cm\]
From the mirror formula, we have $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ , where f is the focal length of the mirror.
Putting the values of u and v in the above equation, we get
$\dfrac{1}{{ - 80}} + \dfrac{1}{{ - 20}} = \dfrac{1}{f}\\
\Rightarrow \dfrac{1}{f} = \dfrac{{1 + 4}}{{ - 80}}\\
\therefore f = - 16\,cm$

Hence the focal length of the given converging mirror is 16 cm.

Note:The distance on the left hand side of the principal axis is taken as positive and on the right hand side of the principal axis is taken as negative. The height of the object is always taken as positive and the height of the inverted image is taken as negative. Focal length of a converging mirror is taken negative.