Question
17. In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phase difference (in degrees) between the waves from the two slits arriving at a point 5 mm from the central maximum when the wavelength is 400 nm? (Convert your result so the angle is between 0 and 3600.) ANS-180°
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Answer #1

Given is:-

Distance between the slits  d = 0.2mm = 0.2 × 1-3m  

Distance between the screen and slits is  D- 100cm 1m

Distance of point from central maxima  5mm = 5 × 10-3m

λ 400nm 400 10-9772

Now,

Path difference - dsine where θ is the angle between the line joining mid point of slits to any pt on the screen and the line joining mid pt of the slits to the central maxima

thus

5 × 1.5×10 3

since tan theta is so small so we can take   tand = sine

thus

path difference -0.2 x 10-3 x 0.005- 10-6

now,

phase difference 2π ー× (path difference)

Delta phi = (rac{2pi}{lambda}) imes (Delta p)

by plugging all the values we get

Delta phi = 5 pi

oxed{Delta phi =180 ^circ}

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