Question

Monochromatic light (vacuum = 612 nm) shines on a soap film (n = 1.33) that has air on either side of it. The light strikes the film perpendicularly. What is the minimum thickness of the film for which constructive interference causes it to look bright in reflected light?

Answer 1

Answer:

t  =   110.079 nm

Explanation:-

we know that

for constructive interference in thin film

2nt = (m + (1/2))λ

where, n is refractive index

for minimum thickness m =0

2nt = λ /2

thickness   t   = λ / 4n

Here  

wavelength of light, λ = 612*10^{-9} m

refractive indices = 1.393$t = \frac{(612*10^{-9})}{(4*1.33)}$$t = 115.03*10^{-9} m$

t  =   110.03 nm