Question

Two narrow, parallel slits separated by 0.850 mm are illuminated by 600-nm light, and the viewing screen is 2.80 m away from the slits. (a) What is the phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe? (b) What is the ratio of the intensity at this point to the intensity at the center of a bright fringe?

Answer 1

a) 7.947 radians

b)
`\mathbf{(I)/(I_(max))=0.4535}`

Step-by-step explanation:

y = Distance from central bright fringe = 2.5 mm

λ = Wavelength = 600 nm

L = Distance between screen and source = 2.8 m

d = Slit distance = 0.85 mm

`tan\theta =(y)/(L)\\\Rightarrow tan\theta =(2.5)/(2800)=0.000892\\\Rightarrow \theta=tan^(-1)0.000892=0.05115^(\circ)`

`\Delta r= dsin\theta\Rightarrow \Delta r=dsin0.05115=7.589*10^(-7)`

a) Phase difference

`\phi=(2\pi)/(\lambda)\Delta r\\\Rightarrow \phi=(2\pi)/(600* 10^(-9))7.589*10^(-7)=7.947\ rad`

∴ Phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe is 7.947 radians

b)
`(I)/(I_(max))=cos^2(\phi)/(2)\\\Rightarrow (I)/(I_(max))=cos^2(7.947)/(2)\\\Rightarrow \mathbf{(I)/(I_(max))=0.4535}`

∴ Ratio of the intensity at this point to the intensity at the center of a bright fringe
`\mathbf{(I)/(I_(max))=0.4535}`