Problem 3: Suppose light has its second-order maximum at 36.5 when falling ona diffraction grating that...

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$Problem 3: Suppose light has its second-order maximum at 36.5 when falling ona diffraction grating that has 5000 lines per ce$

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Answer 1

Answer #1

The grating constant is d = 1/500000 = 2 x 10^-6 m^-1

Therefore the wavelength is

_dsin@_2 x 10-6 x sin(36.5) – 5048 V = 594.8 x 10-m = 594.8 nm

wavelength is 594.8 nanometers

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Answer 2

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Problem 3: Suppose light has its second-order maximum at 36.5 when falling ona diffraction grating that...

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Problem 3: Suppose light has its second-order maximum at 36.5 when falling ona diffraction grating that...