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

Answer:

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:

$xsin\theta = n\lambda$

$xsin\theta = n\times 680.0\times 10^{- 9}$           (1)

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

$xsin\theta = (n + 1)\times 510.0\times 10^{- 9}$         (2)

equate eqn (1) and (2):

$n\times 680.0\times 10^{- 9} = (n + 1)\times 510.0\times 10^{- 9}$

$n = \frac{510.0\times 10^{- 9}}{680.0\times 10^{- 9} - 510.0\times 10^{- 9}}$

n = 3