Question

Establish Lens maker's formula
$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

Answer 1

Let any lens is made up of two surfaces AB and CD of radii of curvatures $R_1$ and $R_2$. A point object is placed on principal axis at O. Image of an object O is formed at I due to AB and this image acts as an object for second surface CD and its image produced at $I_1$.
Let distance of object O from the lens is u image $I_1$ is at distance $v_1$, image I is at distance v
From first surface AB image of O get on $I_1$ that way by refraction formula
$\frac{\mu -1}{R_1}=\frac{\mu }{v_1}-\frac{1}{u}$ .......... (i)
From second surface CD image of $I_1$ get on $I$ by the refraction formula
$\frac{\frac{1}{\mu }-1}{R_2}=\frac{\frac{1}{\mu }-1}{v}-\frac{1}{v_1}$ ......... (ii)
$\mu =\frac{1}{u},u=v_1$ for CD surface because light ray passing from denser to rarer medium.
Equation (2) multiply by $\mu$
$\frac{1}{v}-\frac{\mu }{v_1}=-\frac{1-\mu }{R_2}$ ...... (iii)
Add equation 1 + 3
$\frac{1}{v}-\frac{1}{u}=\frac{\mu -1}{R_1}+\frac{1-\mu }{R_2}$ ......... (iv)
$\frac{1}{v}-\frac{1}{u}=\frac{\mu -1}{R_1}-\frac{\mu -1}{R_2}$ if $=\infty ,v=f$
$\frac{1}{f}-\frac{1}{\infty }=\mu -1(\frac{1}{R_1}-\frac{1}{R_2})\frac{1}{\infty }=0$
$\frac{1}{f}=\mu -1(\frac{1}{R_1}-\frac{1}{R_2})$ ...... (v)
By compare eqn. (4) and (5)
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$