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Question

Derive the Lens Maker's formula.


Solution

Consider the situation shown in figure. $$ADBE$$  is a thin lens. An object 0 is placed on its principal axis. The two spherical surfaces of the lens have their centres at $$C_1$$ and $$C_2$$. The optical centre is at $$P$$ and the principal axis cuts the two spherical surfaces at $$D$$ and $$E$$.

As shown in the figure, after the first refraction image is formed at $$O_1$$ and after second refraction the final image is formed at $$I$$. 
As
the lens is thin, the points $$ D, P and E$$  are all close to each other and we may take the origin at $$P$$ for both these refractions.

The general equation for refraction at a spherical surface is
$$\dfrac{\mu_2}{v}-\dfrac {\mu_1}{u} = \dfrac {\mu_2-\mu_1}{R}$$..........(i) 

For the first refraction, the object is at $$O$$, the
image is at $$O_1$$ and the centre of curvature is at $$C_1$$. If
$$u$$, $$v$$, and $$R$$, denote their x-coordinates,
$$\dfrac{\mu_2}{v_1}-\dfrac {\mu_1}{u} = \dfrac {\mu_2-\mu_1}{R_1}$$.......(ii)

For the second refraction at $$AEB$$, the incident rays $$GH$$ and $$DE$$ diverge from $$O_1$$. Thus, $$ O_1 $$ is the object for
this refraction and its x-coordinate is $$v_1$$. The image is
formed at $$I$$ and the centre of curvature is at $$C_2$$. Their
x-coordinates are $$v$$ and $$R$$ respectively. The light goes
from the medium $$\mu_2$$  to medium $$\mu_1$$.

Hence 
$$\dfrac{\mu_1}{v}-\dfrac {\mu_2}{v_2} = \dfrac {\mu_1-\mu_2}{R_2}$$.......(iii)
Adding (ii) and (iii),
$$\dfrac{1}{v}-\dfrac{1}{u}=(\dfrac{\mu_2}{\mu_1}-1)\times (\dfrac{1}{R_1}-\dfrac{1}{R_2})$$ 

If the object $$O$$ is taken far away from the lens, the image is formed close to the focus. Thus, for $$u$$= $$\infty$$ , $$v=f$$

Hence,
$$\dfrac{1}{f} = (\dfrac {\mu_2}{\mu_1}-1)\times(\dfrac{1}{R_1}-\dfrac{1}{R_2})$$ 
This is lens makers formula. 

669857_629510_ans_8c626af49cfb463f9b4126f05b96ba96.png

Physics

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