Question

A thin film is in the air with a wavelength of 615 nm and is 90 nm thick.

What is the minimum index of refraction?

The light at wavelegth= 410 nm is a first order maximum at 12 degrees. What is the angle for the second order maximum at 670 nm?

The central light is 3 m away when every other slit is covered; how far will the first interference max be?

All but 2 adjacent slits of 10 are covered, now how far is the maximum?

Answer 1

Part a)

Part b)

The formula for diffraction maxima is

d*sin? = m?

d = m?/sin?

'd' is the same for both maxima, so

m₁?₁/sin?₁ = m₂?₂/sin?₂

m₁ = 1

?₁ = 410*10^-9 m

?₁ = 12^o; sin 12 = 0.2079

m₂ = 2

?₂ = 670*10^-9 m

?₂ = ?^o

1*410*10^-9 / 0.2079 = 2*670*10^-9/sin?₂

the angle for the second order maximum at 670 nm, is ?₂ = 42.8^o