Young's experiment is performed with light of wavelength 502nm from excited helium atoms. Fringes are measured carefully on a sc
1 answer:
Answer:
d = 1.30 mm
Explanation:
given,
wavelength of the light source (λ)= 502 nm
slits is separated (d) = ?
distance to form interference pattern(D) = 1.35 m
20 th fringe
y = 10.4 mm = 0.0104 m
now,separation between to slits
where y is the fringe width
d = 1.30 mm
separation between to slits is equal to d = 1.30 mm
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Answer:
Among those two medium, light would travel faster in the one with a reflection angle of (when light enters from the air.)
Explanation:
Let denote the speed of light in the first medium. Let
denote the speed of light in the air. Assume that the light entered the boundary at an angle of
to the normal and exited with an angle of
. By Snell's Law, the sine of
and
would be proportional to the speed of light in the corresponding medium. In other words:
.
When light enters a boundary at the critical angle , total internal reflection would happen. It would appear as if the angle of refraction is now
. (in this case,
.)
Substitute this value into the Snell's Law equation:
.
Rearrange to obtain an expression for the speed of light in the first medium:
.
The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.
For ,
is monotonically increasing with respect to
. In other words, for
in that range, the value of
increases as the value of
increases.
Therefore, compared to the medium in this question with , the medium with the larger critical angle
would have a larger
. such that light would travel faster in that medium.