# State one feature by which the phenomenon of interference can be distinguished from that of diffraction. A parallel beam of light of wavelength 600 nm is incident normally on a slit of width ‘ a ’. If the distance between the slits and the screen is 0.8 m and the distance of 2 nd order maximum from the centre of the screen is 15 mm, calculate the width of the slit. Solution: Difference between interference and diffraction: Interference is due to superposition of two distinct waves coming from two coherent sources . Diffraction is produced as a result of superposition of the secondary wavelets coming from different parts of the same wavefront. Numerical: Here, λ = 600 nm = 600 × 10 −19 = 6 × 10 −7 m D = 0.8 m, x = 15 mm = 1.5 × 10 −3 m, n = 2, a = ? Please check something is wrong in solution...

The formula is absolutely correct but there are a couple of mistakes...

for constructive interference

a(x/D) = nλ

or

the slit width would be

a =  (nλD) / x

here

n = 2

λ = 600 nm = 600 X 10-9 m = 6 X 10-7 m

D = 0.8 m

and x = 15 mm = 15 X10-3 m   (not 1.5 X10-3 m)

so,

a = (2 X 6 X 10-7 X 0.8) / 15 X10-3

or

the slit width is

a = 0.64 X 10-4 m = 6.4 X10-5 m

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deffraction is a result of superimposition of rays comig from different point of a single silt where as inter ference is a result of involvment of two silts in deffractuion the central fringe is  brtight & its amplitude is twice the the secondary fringes.

• 3

the 1st formula is wrong....

in place on (n lamda)...it shud b (2n+1)lambda/2....

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