| ### Statement | | ### Statement |
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| $12.1.20.$ | | $12.1.20.$ |
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| A layer of photoemulsion is applied on a mirror metal substrate. When light | | 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 | | falls normally at a distance of 10−5mm from the metal surface, the emulsion |
| becomes blackened. Explain this effect. Determine the wavelength of light | | becomes blackened. Explain this effect. Determine the wavelength of light |
| incident on a metal surface. At what distance from the substrate surface will | | incident on a metal surface. At what distance from the substrate surface will |
| the second layer of blackened emulsion be located? | | the second layer of blackened emulsion be located? |
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| ### Solution | | ### Solution |
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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. | | 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. |
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| Wavelength | | Wavelength |
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| The first darkening (first antinode) occurs at $d_1 = 10^{-5}\ \text{cm}$. Since $d_1 = \lambda/4$: | | The first darkening (first antinode) occurs at $d_1 = 10^{-5}\ \text{cm}$. Since $d_1 = \lambda/4$: |
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| $\boxed{\lambda = 4 \times 10^{-5}\ \text{cm} = 400\ \text{nm}}$. | | $\boxed{\lambda = 4 \times 10^{-5}\ \text{cm} = 400\ \text{nm}}$. |
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| Distance between darkened layers | | Distance between darkened layers |
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| Consecutive antinodes are separated by half a wavelength: | | Consecutive antinodes are separated by half a wavelength: |
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| $\boxed{x = \frac{\lambda}{2} = 2 \times 10^{-5}\ \text{cm}}.$ | | $\boxed{x = \frac{\lambda}{2} = 2 \times 10^{-5}\ \text{cm}}.$ |
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| #### Answer | | #### Answer |
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