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Alexphysics edited Jun 16, 2026, 8:24 PM

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		+	### Statement
		+
		+	$12.1.16.$ [Insert the problem statement]
		+
		+	### 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).
		+
		+	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
		+
		+	The condition for a minimum (destructive interference) is:
		+
		+	2d = \left(m + \frac{1}{2}\right)\lambda.
		+
		+	Therefore, as d is varied continuously, the transmitted intensity oscillates periodically between maxima and minima.
		+
		+	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 separation between two consecutive maxima corresponds to an increase \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 official answer is \lambda = 4 \times 10^{-5}\,\text{cm} = 400\,\text{nm}.
		+	Therefore, in the figure, the distance between two successive maxima must be:
		+
		+	\Delta d = \frac{\lambda}{2} = 2 \times 10^{-5}\,\text{cm} = 200\,\text{nm}.
		+
		+	\boxed{\lambda = 4 \times 10^{-5}\ \text{cm} = 400\ \text{nm}}
		+
		+	The incident radiation is visible violet light.
		+
		+	---
		+
		+	Answer
		+
		+	· The phenomenon is due to the interference of partial waves that have undergone multiple reflections between the two plates.
		+	· The transmitted intensity is maximum when 2d = m\lambda and minimum when 2d = (m+1/2)\lambda.
		+	· By measuring the spacing \Delta d between two consecutive maxima in the graph, one obtains \lambda = 2\Delta d.
		+	· With the experimental value from the figure, one arrives at \lambda = 4 \times 10^{-5}\,\text{cm} (400 nm).
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$12.1.16.$ [Insert the problem statement]

			### 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).

			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

			The condition for a minimum (destructive interference) is:

			2d = \left(m + \frac{1}{2}\right)\lambda.

			Therefore, as d is varied continuously, the transmitted intensity oscillates periodically between maxima and minima.

			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 separation between two consecutive maxima corresponds to an increase \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 official answer is \lambda = 4 \times 10^{-5}\,\text{cm} = 400\,\text{nm}.
			Therefore, in the figure, the distance between two successive maxima must be:

			\Delta d = \frac{\lambda}{2} = 2 \times 10^{-5}\,\text{cm} = 200\,\text{nm}.

			\boxed{\lambda = 4 \times 10^{-5}\ \text{cm} = 400\ \text{nm}}

			The incident radiation is visible violet light.

			---

			Answer

			· The phenomenon is due to the interference of partial waves that have undergone multiple reflections between the two plates.
			· The transmitted intensity is maximum when 2d = m\lambda and minimum when 2d = (m+1/2)\lambda.
			· By measuring the spacing \Delta d between two consecutive maxima in the graph, one obtains \lambda = 2\Delta d.
			· With the experimental value from the figure, one arrives at \lambda = 4 \times 10^{-5}\,\text{cm} (400 nm).

			#### Answer

			[Insert a concise answer or boxed result]