Question

A parallel beam of light of wavelength 560 nm falls on a thin of oil (refractive index = 1.4) What should be the minimum thickness of the film so that it weakly transmits the light

Answer 1

original path difference due to thin of oil=(u-1)t

or,

path difference due to 1st displacement of fringes=(λ)/2

Therefore,

(u-1)t=(λ)/2..........................(1)

given that,

u=1.4 λ=560nm

now,

$(1.4-1)t=560\times {10}^{9/2}$

t=0.0007mm