Question

g How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for a slit width of 0.110 mm, using light of wavelength 582 nm?

Answer 1

2

Step-by-step explanation:

We know that in the Fraunhofer single-slit pattern,

maxima is given by

`a\text{sin}\theta=(2N+1)/(2)\lambda`

Given values

θ=2.12°

slit width a= 0.110 mm.

wavelength λ= 582 nm

Now plugging values to calculate N we get

`0.110*10^(-3)\text{sin}2.12=((2N+1)/(2))582*10^(-9)`

Solving the above equation we get

we N= 2.313≅ 2