Question

I

I know that the thickness = lambda/(2nsin 30) BUT why? I've also attached some notes I have!

(Please explain using diagram and words for me to rate the answer)

Thin-Film Interference When light is incident on a thin transparent film, the light waves reflected from the front and back surfaces interfere. For near-normal incidence, the wavelength con ditions for maximum and minimum intensity of the light reflected from a film in air are 2L = (m + 2). for m 0, 1,2,... 1n2 brie 2L m for m 0, 1,2,.. where n, is the index of refraction of the film, L is its thickness and (maxima-bright film in air), (35-36) (minima -dark film in air), (35-37) λ is the wavelength of the light in air. If the light incident at an interface between media with dif- ferent indexes of refraction is in the medium with the smaller index of refraction, the reflection causes a phase change of π rad, or half a wavelength, in the reflected wave. Otherwise, there is no phase change due to the reflection. Refraction causes no phase shift.

Answer 1

From the given data,mirror M, shifted in the fringe pattern for a one wavelength 1.0 fringe larger than fringe with other wavel ength m1-m2 = 1 2L 2L 2L 588.9950×10-9 m = 0.2906mm 0.291mm fn 589.5924 x 10-9 m fn