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.

n = 3

Explanation:

We know that the nth-order maximum for the Young's two-slit experiment is given by the following expression:

dsin(θ) = nλ

If two maxima concides we must have the relation:

nλ1 = mλ2  --- eq1

In this particular experiment

m = n+1

λ1 = 680 nm

λ2 = 510 nm

Replacing m in eq1 we have

nλ1 = (n+1)λ2

Solving for n we have:

n = λ2/(λ1-λ2)

Therefore:

n = 510/(680-510)=3

a

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