p37_044 - 44. At the point on the screen where we find the...

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Unformatted text preview: 44. At the point on the screen where we find the inner edge of the hole, we have tan θ = 5.0 cm/30 cm, which gives θ = 9.46◦ . We note that d for the grating is equal to 1.0 mm/350 = 1.0 × 106 nm/350. From mλ = d sin θ, we find 1.0×106 nm (0.1644) 350 d sin θ 470 nm m= = = . λ λ λ Since for white light λ > 400 nm, the only integer m allowed here is m = 1. Thus, at one edge of the hole, λ = 470 nm. However, at the other edge, we have tan θ = 6.0 cm/30 cm, which gives θ = 11.31◦ . This leads to 1.0 × 106 nm λ = d sin θ = sin 11.31◦ = 560 nm . 350 Consequently, the range of wavelength is from 470 to 560 nm. ...
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