How far the first bright diffraction fringe is from the strong central maximum if the screen is 12.0 m away. Express your answer to three significant figures and include the appropriate units.

λ=wavelength = 620 nm = 620 x 10-9 m

d = slit width = 3.40 x 10-3 mm = 3.40 x 10-3 x 10-3 m = 3.40 x 10-6 m       (since 1 mm = 10-3 )

n = order = 1

D = screen distance = 12 m

Using the formula ::

d Sin\theta = (2n + 1) (\lambda/2)

for small angles, Sin\theta = tan\theta = x/D

x = distance of fringe from central maxima

D = distance of screen = 12 m

d (X/D) = (2n + 1) (\lambda/2)

(3.40 x 10-6) (X/12) = (2 + 1) ((620 x 10-9)/2)

X = 3.28 m

Explanation:

The distance of any bright fringe from the central maximum on the screen in a single slit diffraction is given by,

Ym = (2m + 1) lL/2a

where m = 1

l = wave length,

L = width of the slit

a = distance from the centre of the slit

m = 1, l = 640 x 10-9nm , L = 3.4 x 10 -3 and a = 12m.

Ym = (2x1 +1) 640 x 10-9 x 3.4 x 10-3 / 2 x 12

= 2.72 x 10-9m

b.transmitters send radio waves, and receivers capture radio waves.

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