Question

A 0.50-mm-diameter hole is illuminated by light of wavelength 510 nm . What is the width...

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?

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Answer #1

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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