| @@ -1,60 +1,28 @@ |
| | | ### Statement |
| | | |
| | − | $12.1.16.$ |
| | + | $12.1.16.$ 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 radiation propagates perpendicular to the plates. |
| | | |
| | − | 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 |
| | + | <b>Physical phenomenon:</b> |
| | + | We are observing multiple-beam interference (as in a Fabry-Perot interferometer). The incident plane wave partially passes through the first plate, reaches the second, where it again partially passes and partially reflects back. The reflected wave returns to the first plate, reflects from it, and goes to the second one again. At the exit of the system, multiple waves emerge having traveled different optical paths. |
| | + | The path difference between the directly transmitted wave and the wave that underwent two internal reflections is $\Delta = 2d$ (round trip between the plates). |
| | + | When the distance $d$ changes, the path difference changes, leading to a periodic alternation of constructive (maxima) and destructive (minima) interference. |
| | | |
| | − | When a monochromatic plane wave strikes the first semi‑transparent plate perpendicularly: |
| | + | <b>Determining the wavelength from the graph:</b> |
| | + | Condition for a transmission minimum: $2d = \left(m + \frac{1}{2}\right)\lambda$. |
| | + | Condition for the adjacent transmission maximum: $2d = (m + 1)\lambda$. |
| | + | Consequently, when transitioning from an intensity minimum to the adjacent maximum, the optical path difference $2d$ changes by $\lambda/2$. It follows that the distance $d$ itself changes by an amount $\Delta d = \lambda/4$. |
| | | |
| | − | · 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). |
| | + | Let's carefully examine the graph provided in the problem. |
| | + | On the x-axis, the distance between these points is: |
| | + | $$\Delta d = (2 - 1) \cdot 10^{-5}\text{ cm} = 10^{-5}\text{ cm}$$ |
| | + | Since this is the distance from a minimum to the adjacent maximum, we equate it to $\lambda/4$: |
| | + | $$\frac{\lambda}{4} = 10^{-5}\text{ cm} \implies \lambda = 4 \cdot 10^{-5}\text{ cm}$$ |
| | + | (which corresponds to $400\text{ nm}$ — the boundary of visible violet light). |
| | | |
| | − | 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 |
| | − | |
| | − | 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 |
| | − | $\boxed{\lambda = 4 \times 10^{-5}\,\text{cm} (400 nm)}$ |
| | + | The phenomenon is explained by the interference of rays multiply reflected between the plates. |
| | + | $\lambda = 4 \cdot 10^{-5}\text{ cm} = 400\text{ nm}$. |
