A
lens is coated with thin film of
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Solution: Click For PDF Version We will consider interference of reflected rays one from the front surface of the coating and the other that is reflected from the backside of the coating and is refracted back into the incident medium, which is air. As the refractive index of the coating is more than that of air, there is a phase change of
on reflection. Also, the transmitted ray in the coating on
reflection from the lens undergoes a phase
change of
,
as the refractive index of glass the material of the lens is more
than that of the coating (see data shown in the figure). Therefore,
both rays undergo equal amount of additional phase changes. Thee
phase difference between them arises because of the additional
optical path travelled by the transmitted ray inside the coating.In the above context the condition for destructive interference is
where d
is the thickness of the coating. It is given that when a
monochromatic light of wavelength 550 nm is incident normally there
is zero reflection. We use this information for finding the
thickness of the coating. We note that a solution for zero
reflection is that with
We thus find that
If
We next calculate the phase difference for two cases of monochromatic light of wavelengths 450 nm and 650 nm.
(a)
The phase difference will be
and the factor by which the reflection is diminished is
(b)
The phase difference will be
and the factor by which the reflection is diminished is
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to reduce the reflection from the glass
surface. There is zero
reflection for light of wavelength 550 nm at
normal incidence. We
have to find the factor by which the reflection is diminished at
(a) 450 nm and (b) 650
nm.
the two reflected waves have a phase difference of
is the phase difference between the waves that interfere, the
intensity of the resultant is proportional to
.
,
,
