Question

Blue light with a wavelength of 450 nm in air strikes a piece of glass with index of refraction n = 1.50. There is air on both sides of the glass. The following questions refer to this situation.

1). What must be the minimum thickness of the film in order for destructive interference to occur?

a. 75 nm;

b. 150 nm;

c. 225 nm;

d. None of the above.

2). Which one of the rays undergoes a phase jump?

a. Neither ray;

b. Only the ray reflecting on the top surface of the glass;

c. Only the ray reflecting on the bottom surface of the glass;

d. Both rays.

3). If this film (with the thickness found in question 11) is now illuminated with white light, it will appear:

a. Black;

b. Blue;

c. Yellow;

d. White.

Please provide full explanations to each question!

Thank you very much! :)

Answer 1

phase shift in reflecting wave from upper surface = λ/2 ( air to galss . rare to dense so phase shift occurs )

phase shift in reflecting wave from uplower per surface = 2nt

net path difference = Δ = 2nt-λ/2  

for destruction interference  Δ = m*λ/2 so

2nt-λ/2 = λ/2

2nt= λ/2+λ/2

2nt = m*λ for minimum thickness m= 1

t = λ/2n = 450/ (2*1.5) =150 nm

b. Only the ray reflecting on the top surface of the glass;

because it is going from rare air medium to denser glass medium

3)

not blue because it will not reflect from it

2nt = m*λ

λ = 2*1.5*150/m

m1 =450 nm

m2 =225 nm

m3 =150 nm

m4 =112.5 nm

so all these wavelengths will not appear

but these will appear because they have wavelength above 450 nm

Red ~ 700–635 nm ~ 430–480 THz

Orange ~ 635–590 nm ~ 480–510 THz

Yellow ~ 590–560 nm ~ 510–540 THz   ur answer

Green ~ 560–520 nm ~ 540–580 THz