A uniform soap film with index of refraction of 1.38 stretches between two glass

Question

A uniform soap film with index of refraction of 1.38 stretches between two glass plates. A broad spectrum light source shines on the film. The film reflects 597 nm light at normal incidence in the gap between the plates. Ignore the effects of gravity on the film. What is the soap film thickness if it is as thin as possible? What is the smallest wavelength of light reflected from the film coating glass having an index of refraction of 1.65 at normal incidence? What region is this in (UV, visible, IR)? A ray from the source makes an angle of 56 degree with the normal where the soap film lies on the glass. Find the angles of refraction in the soap film and the glass plate.

Explanation / Answer

(a)
T(minimum) = lambda/4
Where lambda is the Wavelength in the soap Film,
= 597/1.38 nm
= 432.6 nm

T = 432.6/4
T = 108.15 nm
Soap Film Thickness, T = 108.15 nm

(b)
Smallest Wavelngth reflected is giveb by,
2 nfilm t = (m+½), for m = 0
2 * 1.65 *108.15 = 1/2*
=  713.8 nm
It comes under, IR Spectrum

(c)
Using Snell's Law -
Angle of refraction in the soap film,
1 * sin(56) = 1.38 * sin(r)
r = 36.9o

Angle of refraction in the Glass Plate,
1.38 * sin(36.9) = 1.33 * sin(r)
r = 38.5 o

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