The distance between the slit and the screen.

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Chapter 38, Problem 1P

(a)

To determine

The distance between the slit and the screen.

(b)

To determine

The width of central maximum.

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Light of wavelength 588.2 nm illuminates a slit of width 0.63 mm. (a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.86 mm from the central maximum? (b) Calculate the width of the central maximum. Step 1 (a) As shown in the figure, dark bands or minima occur where sin 0 = m(2/a). For the first minimum, m = 1 and the distance from the center of the central maximum to the first minimum is y₁ = L tan 8, where L is the distance of the viewing screen from the slit. 32 sin dark = 22/a 31 sin dark = λ/a HE 0 -1 sin dark = -λ/a -2 sin dark = -22/a Viewing screen a Because is very small, we can use the approximation tan sin 0 = m(2/a). Substituting the approximation and solving for the distance to the screen, we have 6.3 x 10 m ³ m ) (₁ L = = y ₁ ( ² ) = x 10-3 m x 10-⁹ m m.

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Light of wavelength 586.5 nm illuminates a slit of width 0.78 mm. (a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.95 mm from the central maximum? m (b) Calculate the width of the central maximum. mm