Answer is incorrect Question 7 of 13 Incorrect Map sapling leaming Light of wave
ID: 252009 • Letter: A
Question
Answer is incorrect Question 7 of 13 Incorrect Map sapling leaming Light of wavelength 638.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 63.5 cm from the slit. The distance on the screen between the second order minimum and the central maximum is 1.29 cm. What is the width of the slit? Number | 6.281 x 10-5 | mim pr Previous Give Up & View Solution e Check Answer 0 Next at a Exit-yo Hint Since the distance between the minimum and the maximum is much smaller than the s angle approximation can be used. the minimum and the maximum is much smaller than the slit-screen distance, the small hisExplanation / Answer
The single slit diffraction pattern is given by
I (angle) = Io sinc^2 (d sin(angle) / lambda)
where the angle is given by
sin(angle) = x / L
where L is the distance from the slit to the observation screen and x is a position in the observation screen referenced at the maximum intensity position, that is I(x=0) = Io. Thus, the intensity distribution is given by
I (x) = Io sinc^2 (d x/L / lambda)
The first zero of the sinc function intensity distribution occurs at
d xo/L / lambda = 1
thus, xo is
d = lambda L / xo = lambda L / D/2
where D is the width of the central maximum. Then
d =0.000000638 * 0.635 / 0.0129 / 2 = 15.70 x 10-6 m = 15.7 um
or 15.70 x 10-3 mm ..............Ans.