Question

An object is placed (A)$20 cm$ , (B)$4 cm$, in front of a Concave Mirror of focal Length $12 cm$. Find the nature and position of the image formed in each case.

Answer 1

Hint: In the question, object distance and focal length of the concave mirror is given. The focal length is negative for the concave mirror. We can find the image distance from the mirror formula and the nature of the image can be found from the magnification formula. If magnification is negative, the image is real and inverted. If magnification is positive, the image is virtual and erect.

Complete step-by-step solution: -
Case1: Distance of object from the mirror, $u = -20 cm$
Focal length of the concave mirror, $f = -12 cm$
Using the mirror formula,
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Put the f and u, we get the values of image distance.
$\dfrac{1}{-12} = \dfrac{1}{v} + \dfrac{1}{-20}$
$\implies \dfrac{1}{20} - \dfrac{1}{12} = \dfrac{1}{v}$
$\implies \dfrac{3-5}{60} = \dfrac{1}{v}$
$\implies v = \dfrac{-60}{2} = -30 cm$
Magnification (m) $= \dfrac{-v}{u}$
$m = - \dfrac{(-30)}{(-20)} = -1.5$
-ve sign shows that the image is real, inverted and enlarged.
 Case2: Distance of object from the mirror, $u = -4cm$
Focal length of the concave mirror, $ f = -12 cm$
Using the mirror formula,
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Put the f and u, we get the values of image distance.
$\dfrac{1}{-12} = \dfrac{1}{v} + \dfrac{1}{-4}$
$\implies \dfrac{1}{4} - \dfrac{1}{12} = \dfrac{1}{v}$
$\implies \dfrac{3-1}{12} = \dfrac{1}{v}$
$\implies v = \dfrac{12}{2} = 6 cm$
Magnification (m) $= \dfrac{-v}{u}$
$m = - \dfrac{(6)}{(-4)} = 1.5$
$+ve$ sign shows that image is virtual and erect.

Note: Magnification is defined as the amount to which the image seems bigger or shorter than the object size. It is shown as the ratio of the image height to the ratio of the object's height. Magnification is the means of enlarging the visible size, not a physical dimension, of something.