A 0.50-mm-diameter hole is illuminated by light of wavelength 510 nm .
What is the width of the central maximum on a screen 2.2 m behind the slit?
Condition for the minimum intenisty in the single slit diffraction pattern is
d sin theta = mL
where d is width of the slit and L is wavelength
here for 1st order m = 1
so
for small angles tan theta = sin theta = theta
also fringe widht is tan theta = Y/D
sin theta = Y/D or mL/d = Y/D
Y = mLD/d
for width on both sides
2Y = 2 * mLD/d
2Y = 2* 1 * 510 e -9 * 2.2/(0.5 e -3)
2Y = 4.48 mm
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