Question

In a Young's two-slit experiment it is found that an nth-order maximum for a wavelength of 680.0 nm coincides with the (n+1)th maximum of light of wavelength 510.0nm. Determine n.

Answer 1

n = 3

Solution:

Since, the slit used is same and hence slit distance 'x' will also be same.

Also, the wavelengths coincide,
`\theta` will also be same.

Using Bragg's eqn for both the wavelengths:

`xsin\theta = n\lambda`

`xsin\theta = n* 680.0* 10^(- 9)` (1)

`xsin\theta = (n + 1)\lambda`

`xsin\theta = (n + 1)* 510.0* 10^(- 9)` (2)

equate eqn (1) and (2):

`n* 680.0* 10^(- 9) = (n + 1)* 510.0* 10^(- 9)`

`n = (510.0* 10^(- 9))/(680.0* 10^(- 9) - 510.0* 10^(- 9))`

n = 3