Chapter 27: Q82PE (page 1000)
Repeat Exercise\({\rm{27}}{\rm{.71}}\), but take the light to be incident at a \({\rm{4}}{{\rm{5}}^ \circ }\)angle.
Short Answer
The wavelength of reflected light is\({\rm{952\;nm}}\)and it will produce black color.
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Chapter 27: Q82PE (page 1000)
Repeat Exercise\({\rm{27}}{\rm{.71}}\), but take the light to be incident at a \({\rm{4}}{{\rm{5}}^ \circ }\)angle.
The wavelength of reflected light is\({\rm{952\;nm}}\)and it will produce black color.
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It is possible that there is no minimum in the interference pattern of a single slit. Explain why. Is the same true of double slits and diffraction gratings?
Calculate the wavelength of light that has its second-order maximum at when falling on a diffraction grating that has lines per centimeter.
What is the wavelength of light falling on double slits separated by 2.00 µm if the third-order maximum is at an angle of 60.0º ?
An inventor notices that a soap bubble is dark at its thinnest and realizes that destructive interference is taking place for all wavelengths. How could she use this knowledge to make a non-reflective coating for lenses that is effective at all wavelengths? That is, what limits would there be on the index of refraction and thickness of the coating? How might this be impractical?
Red light of wavelength of 700 nm falls on a double slit separated by 400 nm. (a) At what angle is the first-order maximum in the diffraction pattern? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
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