Thin films, including soap bubbles and oil show patterns of alternative dark and bright regions resulting from interference among the reflected ligth waves. If two waves are in phase, their crests and troughs will coincide. The interference will be cosntructive and the amlitude of resultant wave will be greater then either of constituent waves. If the two wave are not of phase by half a wavelength `(180^(@))`, the crests of one wave will coincide width the troughs of the other wave. The interference will be destructive and the ampliutde of the resultant wave will be less than that of either consituent wave.
1. When incident light I, reaches the surface at point a, some of the ligth is reflected as ray `R_(a)` and some is refracted following the path ab to the back of the film.
2. At point b, some of the light is refracted out of the film and part is reflected back through the film along path bc. At point c, some of the light is reflected back into the film and part is reflected out of the film as ray `R_(c)`.

`R_(a)` and `R_(c)` are parallel. However, `R_(c)` has travelled the extra distance within the film fo abc. If the angle of incidence is small, then abc is approxmately twice the film's thickness .
If `R_(a)` and `R_(c)` are in phase, they will undergo constructive interference and the region ac will be bright. If `R_(a)` and `R_(c)` are out of phase, they will undergo destructive interference and the region ac will be dark.
I. Refraction at an interface never changes the phase of the wave.
II. For reflection at the interfere between two media 1 and 2, if `n_(1) gt n_(2)`, the reflected wave will change phase. If `n_(1) lt n_(2)`, the reflected wave will not undergo a phase change.
For reference, `n_(air) = 1.00`.

III. If the waves are in phase after reflection at all intensities, then the effects of path length in the film are:
Constrictive interference occurs when `2 t = m lambda // n, m = 0, 1,2,3`,...
Destrcutive interference occurs when `2 t = (m + (1)/(2)) (lambda)/(n)`,
`m = 0, 1, 2, 3`,...
If the waves are `180^(@)` out of the phase after reflection at all interference, then the effects of path length in the film ara:
Constructive interference occurs when
`2 t = (m + (1)/(2)) (lambda)/(n), m = 0, 1, 2, 3`,...
Destructive interference occurs when
`2 t = (m lambda)/(n) , m = 0, 1, 2, 3`,...
The average human eye sees colors with wavelength between 430 nm to 680 nm. For what visible wavelength (s) will a 350 nm thick `(n = 1.35)` soap film produce maximum destructive interference?

Options

A. 945 nm
B. 473 nm
C. 315 nm
D. None of these

Answer

b

Explanation

Choices (a) and (c ) are eliminated immediately since are outside the visible range. For soap, `n_("air") gt n_("soap")`. The reflected wave, ray a, undergoes a phase change at the air-soap water interface. Ray `R_(b)` does not change phase at the soapy water-air interface. Based on interface reflections, the two rays are out of phase. To maintain this, the optical path difference, 2t, must not produce a phase change in `R_(b)`. Therefore, the path must be an integer multiple of `lambda`
`2 t = m lambda//n implies lambda = 2 t n//m = 2(35nm) (1.35)//m = 945// m`
For `m = 1, lambda = 945 nm`
For `m = 2, lambda = 473 nm` [this is choice (b)]
For `m = 3, lambda = 315 nm`
Choice (b), `lambda = 473 nm`, is the only choice in the visible range.