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| | | ### Statement |
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| | − | $12.1.20.$ |
| | + | $12.1.20$. A layer of photoemulsion is applied on a mirror metal substrate. When light falls normally at a distance of $10^{-5}\text{ mm}$ from the metal surface, the emulsion becomes blackened. Explain this effect. Determine the wavelength of light incident on the metal surface. At what distance from the substrate surface will the second layer of blackened emulsion be located? |
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| | − | A layer of photoemulsion is applied on a mirror metal substrate. When light |
| | − | falls normally at a distance of 10−5mm from the metal surface, the emulsion |
| | − | becomes blackened. Explain this effect. Determine the wavelength of light |
| | − | incident on a metal surface. At what distance from the substrate surface will |
| | − | the second layer of blackened emulsion be located? |
| | + |  |
| | | |
| | | ### Solution |
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| | + | Light incident normally on the metallic mirror reflects and interferes with the incident light, generating a standing electromagnetic wave. As shown in the previous problems, the electric field vanishes at the surface of an ideal metal—a <b>node</b> of the electric field ($E = 0$) is formed there. |
| | + | The <b>antinodes</b> (maxima of electric field intensity) are located at distances of $\lambda/4$, $3\lambda/4$, $5\lambda/4$, etc., from the surface. The photographic emulsion is sensitive specifically to the electric component of the field, so its darkening occurs at the locations of these antinodes. |
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| | − | Light incident normally on the metallic mirror interferes with the reflected light, generating a standing wave. The electric field has a node at the metal surface $(E = 0) $ and antinodes (intensity maxima) at distances $\lambda/4, 3\lambda/4, 5\lambda/4, \dots $from the surface. The photographic emulsion darkens at those antinodes. |
| | + | <b>Wavelength:</b> |
| | + | The first darkening (first antinode) occurs at a distance $d_1 = 10^{-5}\text{ mm}$. Since $d_1 = \lambda/4$: |
| | + | $$\frac{\lambda}{4} = 10^{-5}\text{ mm} \implies \lambda = 4 \cdot 10^{-5}\text{ mm}$$ |
| | + | <i>(For reference: this is $4 \cdot 10^{-8}\text{ m} = 40\text{ nm}$, which corresponds to ultraviolet radiation).</i> |
| | | |
| | − | Wavelength |
| | + | <b>Distance to the second layer:</b> |
| | + | The second layer of blackened emulsion forms at the location of the second antinode of the standing wave. The distance from the metal surface to the second antinode is: |
| | + | $$d_2 = \frac{3\lambda}{4} = 3 \cdot d_1 = 3 \cdot 10^{-5}\text{ mm}$$ |
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| | − | The first darkening (first antinode) occurs at $d_1 = 10^{-5}\ \text{cm}$. Since $d_1 = \lambda/4$: |
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| | − | $\boxed{\lambda = 4 \times 10^{-5}\ \text{cm} = 400\ \text{nm}}$. |
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| | − | Distance between darkened layers |
| | − | |
| | − | Consecutive antinodes are separated by half a wavelength: |
| | − | |
| | − | $\boxed{x = \frac{\lambda}{2} = 2 \times 10^{-5}\ \text{cm}}.$ |
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| | | #### Answer |
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| | − | |
| | − | $\boxed{\lambda = 4 \times 10^{-5}\ \text{cm} = 400\ \text{nm}}$. |
| | − | |
| | − | $\boxed{x = \frac{\lambda}{2} = 2 \times 10^{-5}\ \text{cm}}.$ |
| | + | The effect is due to the formation of a standing wave and the location of the electric field antinodes. |
| | + | $\lambda = 4 \cdot 10^{-5}\text{ mm}$; the second layer is at a distance of $3 \cdot 10^{-5}\text{ mm}$. |
