Question

The thickness of soap film (n=1.46) is 300nm.what wavelength of light in the visible region can be used to constructively reflect off the film?

Answer 1

for constructive interference ,
$\mu t =\frac{(2n+1)\lambda}{4}$
where $\mu$ is refractive index of soap film, $\lambda$ wavelength of light sectrum and t is thickness of soap film.

Here, t = 300nm , $\mu$ = n = 1.46

so, 1.46 × 300 = (2n + 1)λ/4

=> 1.46 × 300 × 4 = (2n + 1)λ

=> 1.46 × 1200/(2n + 1) = λ

for n = 1 , λ = 1.46 × 1200/3 = 1.46 × 400
λ = 5.84 × 100 = 584 nm

for n = 2, λ = 1.46 × 1200/5 = 1.46 × 240
λ = 350.4 nm

but we know, visible region is 390 - 700 nm
so, λ = 584 nm ✓ and λ ≠ 350.4 nm

hence, answer is 584 nm