When two parallel translucent mirror plates are extended, the intensity of
+
electromagnetic radiation transmitted through these plates periodically changes
+
depending on the distance between them. Explain this phenomenon and use
+
the figure to determine the wavelength of the incident radiation. The radia-
+
tion propagates perpendicular to the plates.
+
### Solution
Physical phenomenon: interference due to multiple reflections
When a monochromatic plane wave strikes the first semi‑transparent plate perpendicularly:
· Part of it is transmitted, travels to the second plate, where again part is reflected and part is transmitted.
· The wave reflected at the second plate returns to the first, is partially reflected there, and so on.
· The waves that finally emerge through the second plate (transmitted) are many, each having traveled a different optical path.
The path difference between two successive transmitted waves is 2d (back and forth between the plates).
@@ -15,40 +21,40 @@Solution
These waves interfere. The condition for constructive interference (maximum transmission) is that the path difference be an integer multiple of the wavelength in the medium (assuming air or vacuum, index n = 1):
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2d = m\lambda, \qquad m = 1, 2, 3, \dots
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$2d = m\lambda, \qquad m = 1, 2, 3, \dots$
The condition for a minimum (destructive interference) is:
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2d = \left(m + \frac{1}{2}\right)\lambda.
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$2d = \left(m + \frac{1}{2}\right)\lambda$
Therefore, as d is varied continuously, the transmitted intensity oscillates periodically between maxima and minima.
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Determination of \lambda from the graph
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Determination of $\lambda$ from the graph
The figure in the problem shows the transmitted intensity I as a function of the distance d between the plates (or sometimes as a function of time if d changes linearly with it). In that graph, equally spaced maxima are observed.
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· The separation between two consecutive maxima corresponds to an increase \Delta d that satisfies:
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2\Delta d = \lambda\quad\Rightarrow\quad\lambda = 2\,\Delta d.
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· The separation between two consecutive maxima corresponds to an increase $\Delta d $
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that satisfies:
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$2\Delta d = \lambda\quad\Rightarrow\quad\lambda = 2\,\Delta d$
· The separation between a maximum and the next minimum (or vice versa) corresponds to \lambda/4.
−
The official answer is \lambda = 4 \times 10^{-5}\,\text{cm} = 400\,\text{nm}.
When two parallel translucent mirror plates are extended, the intensity of
electromagnetic radiation transmitted through these plates periodically changes
depending on the distance between them. Explain this phenomenon and use
the figure to determine the wavelength of the incident radiation. The radia-
tion propagates perpendicular to the plates.
### Solution
### Solution
Physical phenomenon: interference due to multiple reflections
Physical phenomenon: interference due to multiple reflections
When a monochromatic plane wave strikes the first semi‑transparent plate perpendicularly:
When a monochromatic plane wave strikes the first semi‑transparent plate perpendicularly:
· Part of it is transmitted, travels to the second plate, where again part is reflected and part is transmitted.
· Part of it is transmitted, travels to the second plate, where again part is reflected and part is transmitted.
· The wave reflected at the second plate returns to the first, is partially reflected there, and so on.
· The wave reflected at the second plate returns to the first, is partially reflected there, and so on.
· The waves that finally emerge through the second plate (transmitted) are many, each having traveled a different optical path.
· The waves that finally emerge through the second plate (transmitted) are many, each having traveled a different optical path.
The path difference between two successive transmitted waves is 2d (back and forth between the plates).
The path difference between two successive transmitted waves is 2d (back and forth between the plates).
@@ -15,40 +21,40 @@Solution
These waves interfere. The condition for constructive interference (maximum transmission) is that the path difference be an integer multiple of the wavelength in the medium (assuming air or vacuum, index n = 1):
These waves interfere. The condition for constructive interference (maximum transmission) is that the path difference be an integer multiple of the wavelength in the medium (assuming air or vacuum, index n = 1):
2d = m\lambda, \qquad m = 1, 2, 3, \dots
$2d = m\lambda, \qquad m = 1, 2, 3, \dots$
The condition for a minimum (destructive interference) is:
The condition for a minimum (destructive interference) is:
2d = \left(m + \frac{1}{2}\right)\lambda.
$2d = \left(m + \frac{1}{2}\right)\lambda$
Therefore, as d is varied continuously, the transmitted intensity oscillates periodically between maxima and minima.
Therefore, as d is varied continuously, the transmitted intensity oscillates periodically between maxima and minima.
Determination of \lambda from the graph
Determination of $\lambda$ from the graph
The figure in the problem shows the transmitted intensity I as a function of the distance d between the plates (or sometimes as a function of time if d changes linearly with it). In that graph, equally spaced maxima are observed.
The figure in the problem shows the transmitted intensity I as a function of the distance d between the plates (or sometimes as a function of time if d changes linearly with it). In that graph, equally spaced maxima are observed.
· The separation between two consecutive maxima corresponds to an increase \Delta d that satisfies:
· The separation between two consecutive maxima corresponds to an increase $\Delta d $
2\Delta d = \lambda\quad\Rightarrow\quad\lambda = 2\,\Delta d.
that satisfies:
$2\Delta d = \lambda\quad\Rightarrow\quad\lambda = 2\,\Delta d$
· The separation between a maximum and the next minimum (or vice versa) corresponds to \lambda/4.
· The separation between a maximum and the next minimum (or vice versa) corresponds to \lambda/4.
The official answer is \lambda = 4 \times 10^{-5}\,\text{cm} = 400\,\text{nm}.
