Question

Light of wavelength 500 nm falls normally on 50 slits that are 2.5×10^{−3} mm wide and spaced 5.0×10^{−3} mm apart. How many interference fringes lie in the central peak of the diffraction pattern?

Answer 1

The central peak of the diffraction pattern will contain 50 interference fringes.

Step-by-step explanation:

When light of wavelength λ passes through a diffraction grating with N slits, the angles θ of the interference fringes are given by the equation ( sin θ = mλ / d ), where m is the order of the fringe, λ is the wavelength, and d is the separation between slits.

In this case, the light has a wavelength of 500 nm, the grating consists of 50 slits, and the separation between slits is 5.0×1
`0^((-3))` mm. We can use the equation mentioned earlier to find the angle θ for the central peak (m=0). Rearranging the formula gives ( m = d \sin θ / λ ). For the central peak, ( sin θ ) is maximum (equal to 1), so ( m = d / λ ).

Now, substitute the values:
`\( m = (5.0×10^(-3) mm) / (500 nm) \)`. The units are consistent, and the result is 10. Therefore, there are 10 interference fringes on either side of the central maximum. Since interference fringes occur on both sides of the central maximum, the total number of interference fringes in the central peak is 2 times 10, which is 20. However, each fringe has two parts (bright and dark), so the number of interference fringes is effectively 20 / 2 = 10. Therefore, the central peak of the diffraction pattern contains 10 interference fringes, and since there are 50 slits, the total number of interference fringes is 50.