Question

In a Young's double-slit experiment the wavelength of light used is 466 nm (in vacuum), and the separation between the slits is 1.3 × 10^-6 m. Determine the angle that locates: (a) the dark fringe for which m = 0 (b) the bright fringe for which m = 1 (c) the dark fringe for which m = 1 and (d) the bright fringe for which m = 2.

Answer 1

Part a)

`\theta_1 = 10.3 degree`

Part b)

`\theta_2 = 21 degree`

Part c)

`\theta_3 = 32.5 degree`

Part d)

`\theta_4 = 45.8 degree`

Step-by-step explanation:

For the position of dark fringe the path difference of light is odd multiple of half of the wavelength

so here we will have

`dsin\theta = (2m + 1)/(2)\lambda`

part a)

for m = 0 we have

`d sin\theta = (\lambda)/(2)`

`(1.3 * 10^(-6))sin\theta = (466 * 10^(-9))/(2)`

`\theta = 10.3 degree`

Part b)

Similarly for the maximum intensity we will have path difference must be integral multiple of wavelength

so we have

`d sin\theta = N\lambda`

for m = 1 we have

`(1.3 * 10^(-6))sin\theta = 466 * 10^(-9)`

`\theta = 21 degree`

Part c)

for m = 1 we have

`d sin\theta = (3\lambda)/(2)`

`(1.3 * 10^(-6))sin\theta = (3(466 * 10^(-9)))/(2)`

`\theta = 32.5 degree`

Part d)

for m = 2 again we have

`d sin\theta = N\lambda`

`(1.3 * 10^(-6))sin\theta = 2* 466 * 10^(-9)`

`\theta = 45.8 degree`