Final answer:
Using the single-slit diffraction formula, we find the slit width that produces the first minimum for 600-nm light at 60.0° to be approximately 0.6928 μm. The wavelength of light that has its first minimum at 62.0° is calculated to be around 611.5 nm, with the closest given options being 0.50 μm and 600 nm respectively.
Step-by-step explanation:
To solve the problem presented in the question, we'll use the equation for the diffraction minimum condition for a single slit, which is a × sin(θ) = m × λ, where a is the width of the slit, θ is the diffraction angle, m is the order of the minimum (1 for the first minimum), and λ is the wavelength of light.
(a) The width of a single slit that produces its first minimum at 60.0° for 600-nm light can be calculated as follows:
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- a = m × λ / sin(θ)
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- a = 1 × 600 nm / sin(60.0°)
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- a = 1 × 600 nm / 0.8660
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- a ≈ 692.8 nm or 0.6928 μm
(b) Find the wavelength of light that has its first minimum at 62.0°:
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- λ = a × sin(θ) / m
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- λ = 0.6928 μm × sin(62.0°)
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- λ = 0.6928 μm × 0.8829
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- λ ≈ 611.5 nm
Thus, the correct answers are closest to option (c) (a) 0.50 μm (rounding to the nearest option given), and (b) 600 nm.