Light of wavelength 500 nm illuminates a round 0.50-mm diameter hole. A screen is placed 6.0 m behind the slit.. Find the width of the central maximum on the screen. A) 6.0 mm.
B) 12 mm.
C) 260 mm.
D) 7.3 mm.
E) 15 mm.

15mm

Explanation:

We know that for circular holes first dark spot is given by

sin စ = 1.22 λ/D

Also we know that at the same time

tan စ = r/L

So

r = L tanစ = 6 x tan( arcsin(1.22x 500 x10^9/0.50 x 10^ 3))

= 0.0073 m = 7.3 mm

However since the size is twice that so 14.6 mm which is approx 15mm

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