A soap film with index of refraction greater than air is formed on a circular wire frame that is held in a vertical plane. The film is viewed by reflected light from a white-light source. Bands of color are observed at the lower parts of the soap film, but the area near the top appears black. A correct explanation for this phenomenon would involve which of the following?
I. The top of the soap film absorbs all the light incident on it; none is transmitted.
II. The thickness of the top part of the soap film has become much less than a wavelength of visible light.
III. There a phase change of 180o for all wave-lengths of light reflected from the front surface of the soap film.
IV. There is no phase change for any wavelength of light reflected from the back surface of the soap film
A. I only
B. II and III only
C. III and IV only
D. I, II and III
E. II, III and IV
(GR9277 #21)
Solution:
I. FALSE.
Soap film does not absorb light.
II. TRUE.
The gravity effect on water molecules decreases the thickness, t of the top of the soap film, become much less than a wavelength, λ of visible light.
When t ≪ λ, the path difference between the front and the rear surface reflections is small compared to λ. It causes an insignificant phase shift. The result is complete destructive interference of the 2 light waves, creating a black band.
III and IV. TRUE.
Thin film:
n1〈 n2 → there is 180o phase change
n1 〉n2 → no phase change
n1 〉n2 → no phase change
n = refractive index
From air to soap (front surface):
nair〈 nsoap → 180o phase change
From soap to air (back surface):
nsoap 〉nair → no phase change
Answer: E
nsoap 〉nair → no phase change
Answer: E
No comments :
Post a Comment