Question

Derive the lens maker formula in case of double convex lens 3 marks

Answer 1

Answer:

$\sf{\dfrac{1}{f}=\left(\dfrac{\mu_2}{\mu_1}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)}$

Step-by-step Explanation:

Consider figure with the two refracting surfaces having radii of curvature equal to $\sf{R_1}$ and $\sf{R_2}$, respectively.

Now, using the result that we obtained for refraction at single spherical surface, we get

For the first surface, $\sf{\dfrac{\mu_2}{v_1}-\dfrac{\mu_1}{u}=\dfrac{\mu_2-\mu_1}{R_1}.......\left(1\right)}$

For the second surface, $\sf{\dfrac{\mu_1}{v}-\dfrac{\mu_2}{v_1}=\dfrac{\mu_1-\mu_2}{R_2}.......\left(2\right)}$

Adding Eqs. (1) and (2), we get

$\sf{\dfrac{\mu_1}{v}-\dfrac{\mu_1}{u}=\left(\mu_2-\mu_1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)}$

$\sf{mu_1\left(\dfrac{1}{v}-\dfrac{1}{u}\right) =\left(\mu_2-\mu_1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)}$

$\sf{\left(\dfrac{1}{v}-\dfrac{1}{u}\right)=\left(\dfrac{\mu_2}{\mu_1}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)}$

In the above equation, if the object is at infinity and image is formed at the focus, then

$\sf{u=\infty\:and\:v=f}$

Therefore,

$\boxed{\sf{\dfrac{1}{f}=\left(\dfrac{\mu_2}{\mu_1}\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)}}$