Question

In a single slit diffraction experiment first minimum for lambda 1 is equal to 660 nm coincides with first maximum for wavelength lambda to calculate lambda 2

Answer 1

Explanation:

In a single slit diffraction experiment, position of minima is given by dsinθ=nλ

So for first minima of red sinθ=1×( d\λ R )

and as first maxima is midway between first and second minima, for wavelength λ

′

Its position will be

dsinθ = 2λ ′+2λ ⇒sinθ ′=2d3λ ′

According to given condition sinθ=sinθ

′

⇒λ ′= 2/3 λ λR so λ ′= 2/3×6600=440 nm=4400 A˚