# Lateral Magnification

Linear magnification of a Spherical mirror is the ratio of the size of the image to the size of object .

$\displaystyle m = \frac{Size \;of \;image (h_2) }{size \;of \; object (h_1)} = \frac{A’B’}{AB}$

Magnification (m) will be +ve when h2 is +ve i.e image is erect and m will be -ve when h2 is -ve i.e. image is inverted . (Assuming h1 is always +ve)

In new cartesian sign convention, we define magnification in such a way that a negative sign (of m) implies inverted image and vice-versa. A real image is always inverted one and a virtual one is always erect.

Keeping these points in mind and that the real object and its real image would lie on the same sides in case of mirror and on opposite sides in case of lenses, we define m as in case of reflection by spherical mirror as :

$\displaystyle m = -\frac{v}{u}$

Problem : An erect image is three times the size of object is obtained with a concave mirror of radius of curvature 36 cm . What is the position of the object ?

Solution : Since image is erect , m = + 3

$\displaystyle m = -\frac{v}{u}$

$\displaystyle 3 = -\frac{v}{u}$

v = – 3 u

$\displaystyle \frac{1}{v} + \frac{1}{u} = \frac{1}{f} = \frac{2}{R}$

$\displaystyle \frac{1}{- 3 u} + \frac{1}{u} = \frac{1}{f} = \frac{2}{-36}$

$\displaystyle -\frac{1}{3 u} + \frac{1}{u} = – \frac{1}{18}$

$\displaystyle \frac{-1 + 3}{3 u} = – \frac{1}{18}$

$\displaystyle \frac{2}{3 u} = – \frac{1}{18}$

3 u = -36

u = -12 cm

Example : A short linear object of length L lies on the axis of a spherical mirror of focal length f at a distance b from the mirror. Find the size of the image ?

Solution :

$\displaystyle \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

By differentiating both sides

$\displaystyle \frac{dv}{v^2} = – \frac{du}{u^2}$

du : size of object = L

dv : size of image

u : object distance

v : image distance.

$\displaystyle dv = – \frac{v^2}{u^2}(du)$

$\displaystyle \frac{1}{v} = \frac{1}{f} – \frac{1}{u}$

$\displaystyle v = \frac{u f}{u – f}$

$\displaystyle dv = – \frac{u^2 f^2}{(u-f)^2 u^2}du$

$\displaystyle dv = -\frac{f^2}{(b-f)^2}L$

Negative sign implies that object is lying between u and u + du and the image will lie between v and v − dv

Example : A point source S is placed midway between two converging mirrors having equal focal length f as shown in the figure. Find the values of d for which only one image is formed.

Solution : If the image is formed of the object, it must be formed at the center of curvature C.

since only one image is formed by two mirrors, the object should be placed at their common center of curvature as shown in the figure; the mirrors are the part of a sphere of center

⇒ 2f + 2f = d

⇒ d = 4f

Note : If the distance between the mirrors is 2f then also image will be found at the same position. So second value of d = 2f

Exercise : A concave mirror forms a real image on a screen of thrice the linear dimension of the object. Object and screen are moved until image is twice the size of the object. If the shift of the object is 6 cm, find the shift of the screen and the focal length of the mirror.