Question

If instead of 450 nm light, another light of wavelength 680 nm is used, the number of peaks of the interference formed in the central peak of the envelope of the diffraction pattern will be:

• A. \(2\)
• B. \(4\)
• C. \(6\)
• D. \(9\)

Hint:

Using a longer wavelength not only alters the diffraction pattern's angular width but also impacts the density and number of interference fringes within it.

Answer 1

The Correct Option is C

Step 1: Recalculate the angular width of the central maximum with the new wavelength.
With a wavelength of 680 nm, the angular width of the central maximum in the diffraction pattern is recalculated using the formula:
\[ \theta \approx \frac{\lambda}{a} \] where \(\lambda\) is now 680 nm, and \(a\) remains as 2 m. Thus: \[ \theta \approx \frac{680 \times 10^{-9}}{2} = 340 \times 10^{-9} \, radians} \] Step 2: Determine the separation of the interference fringes with the new wavelength.
The fringe separation in the double-slit interference pattern, given by:
\[ \Delta y = \frac{\lambda L}{d} \] will be recalculated with the new \(\lambda = 680 \times 10^{-9}\) m. Assuming \(d = 6\) m, we get:
\[ \Delta y \approx \frac{680 \times 10^{-9} L}{6} \] Step 3: Calculate the number of peaks within the central maximum with the new wavelength.
The total width of the central diffraction peak, \(2\theta\), and the number of interference fringes fitting within this width, are recalculated:
\[ Number of peaks} = \frac{2\theta L}{\Delta y} = \frac{2 \times 340 \times 10^{-9} L}{\frac{680 \times 10^{-9} L}{6}} = 6 \] Given the larger wavelength, the angular width of the central maximum is wider, allowing more interference fringes to fit within.