Mumbai university > FE > SEM 2 > Applied Physics II

Marks: 8M

Year: Dec 2014

Consider a planoconvex lens of radius ‘R’ as shown.

As radius is comparatively large, the space between lens and base can be considered as wedge shaped.

Thus path difference is given by

$Δ = 2μt cos (r_e + α) ± \frac{λ}{2}.....(r_e = angle \ \ of \ \ refraction)$

For almost normal incidence,

$cos (r_e + α) ≈ 1$

$Δ = 2t + \frac{λ}{2}....(i)$

Now,

$R^2 = (R - t)^2 + r^2$

$R^2 = R^2 - 2Rt + t^2 + r^2$

$2t = \frac{t^2}{R} + \frac{r^2}{R}$

$R \gt\gt t^2 , \frac{k^2}{R} ≈ 0$

$2t = \frac{r^2}{R} = \frac{D^2}{4R}.....(ii)$ [D = 2r]

From (i) and (ii)

$Δ = \frac{D^2}{4R} + \frac{λ}{2}.....(iii)$

For Bright Rings:-

$Δ = nλ$

$\frac{D_n^2}{4R} + \frac{λ}{2} = nλ.....[D_n = dia. \ \ of \ \ nth \ \ bright \ \ ring]$

$\frac{D_n^2}{4R} = (2n - 1)\frac{λ}{2}$

$R = \frac{D_n^2}{2(2n - 1)λ}$

Using above formula, by calculating diameter of nth bright ring for a given wavelength of light, we can calculate the radius of curvature.