A light of wavelength $5880A^0$ is incident on a thin film of $μ =1.5$ such that angle of refraction is $60^0$ .

Mumbai university > FE > SEM 2 > Applied Physics II

Marks: 3M

Year: Dec 2014

Given:

$λ = 5.88 × 10^{-7} m$

$μ = 1.5$

$r = 60^0$

To find :

t

Solution :

For Destructive interference on a reflected side of thin film:-

$2 μ cos r = nλ$

(For minimum thickness, n=1)

$2(1.5)t cos(60) = 5.88 × 10^{-7}$

$t = 3.92 × 10^{-7}$

$t = 0.392 μm$