Question

The focal length of a concave mirror is f and the distance from the object to the principal focus is p. The ratio of the size of the real image to the size of the object is:
A. \[ - \dfrac{f}{p}\]
B. ${\left( {\dfrac{f}{p}} \right)^2}$
C. ${\left( {\dfrac{f}{p}} \right)^{\dfrac{1}{2}}}$
D. $ - \dfrac{p}{f}$
E. $ - fp$

Answer 1

Hint: The distance of the object from the mirror u is equal to the sum of focal length f and the distance of the object from the focus p,$u = - (p + f)$. Now we use the mirror formula to find out the distance of the image from the mirror,$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$.
Ratio of size of the image to that of the object is known as the magnification for the mirror, $m = - \dfrac{v}{u}$.

Complete step by step answer:Given the focal length of the mirror is f.
The distance of the object from the focus is p.
The distance of the object from the mirror is $u = - (p + f)$.
By using conventions , distance of object from mirror and focal length of mirror is negative.
Using the mirror formula,
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
$
  \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} \\
 \Rightarrow \dfrac{1}{v} = \dfrac{1}{{ - f}} - \dfrac{1}{ - (p + f) \\ \\ } \\
  \Rightarrow \dfrac{1}{v} = - \dfrac{1}{f} + \dfrac{1}{{p + f}} \\
  \Rightarrow \dfrac{1}{v} = \dfrac{p}{{f(p + f)}} \\
  \therefore v = \dfrac{{f(p + f)}}{p} \\
 $

Now to find out the ratio of the size of the image to that of the object which is also known as the magnification of the mirror, $m = - \dfrac{v}{u}$.
Substituting the value of v and u in the above equation, we get
 $
  m = - \dfrac{{ - \left[ {\dfrac{{f(p + f)}}{p}} \right]}}{{ - (p + f)}} \\
  \therefore m = - \dfrac{f}{p} \\
 $
Therefore the required ratio is $ - \dfrac{f}{p}$.

Note:Conventions for the measurement of the distance from the mirror should be kept into consideration. All the distances should be measured from the pole. The distance in the direction of the incident ray is taken as positive and vice versa.