Magnification

(i) Magnification in Mirror

In new Cartesian sign convention, we define magnification is 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:

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

(ii) Lateral Magnification for Refracting Spherical Surface

$M=\frac{-\left(A^{\prime }{B}^{\prime }\right)}{AB}.Now,\frac{\sin i}{\sin r}=\frac{{\mu }_2}{{\mu }_1}$

$Asi,r\to 0,\Rightarrow \frac{\tan i}{\tan r}=\frac{{\mu }_2}{{\mu }_1}$

$\Rightarrow \frac{{}^{AB}/{}_{PA}}{{}^{A^{\prime }{B}^{\prime }}/{}_{P{A}^{\prime }}}=\frac{{\mu }_2}{{\mu }_1}\Rightarrow -\frac{A^{\prime }{B}^{\prime }}{AB}=\frac{{}^{-P{A}^{\prime }}/{}_{{\mu }_2}}{{}^{PA}/{}_{{\mu }_1}}$

Hence, $m=\frac{{}^v/{}_{{\mu }_2}}{{}^u/{}_{{\mu }_1}}$

(iii) Magnification of A Thin Lens

$M=\frac{v}{u}=\frac{-\left(O{A}^{\prime }\right)}{\left(AO\right)}$

Illustration 1: An object is placed at a distance of 10 cm to the left on the axis of a convex lens A of focal length 20 cm. A second convex lens of focal length 10 cm is placed co-axially to the right of the lens A at a distance of 5 cm from A. Find the position of the final image and its magnification. Trace the path of the rays.

Solution: Here for 1st lens, u₁ = –10 cm

f₁ = 20 cm.

$\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}\Rightarrow \frac{1}{v_1}=\frac{1}{20}-\frac{1}{10}$

$\Rightarrow v_1=-20cm$

i.e. the image is virtual and hence lies on the same side of the object. This will behave as an object for the second lens.

For 2nd lens

$\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}$

Here $u_2=-\left(20+5\right)$

$f_2=10cm$

$\frac{1}{v_2}+\frac{1}{25}=\frac{1}{10}\Rightarrow v_2=\frac{50}{3}=16\frac{2}{3}cm$

i.e. the final image is at a distance of $16\frac{2}{3}cm$ on the right of the second lens.

The magnification of the image is given by,

$m=\frac{v_1}{u_1}\cdot \frac{v_2}{u_2}=\frac{20}{10}\cdot \frac{50}{3\times 25}=\frac{4}{3}=1.33$