Question

in a double - slit diffraction experiment, two slits of width 14.2×10⁻⁶ m are separated by a distance of 31.2×10⁻⁶ m, and the wavelength of the incident light is 651 nm. the diffraction pattern is viewed on a screen 4.43 m from the slits. assume ( i_p ) is the intensity at a point p, a distance ( y = 81.5 ) cm on the screen from the central maximum. which of the following best describes where the point p is on the double - slit interference pattern?

• the point p is between the ( m = 9 ) maximum and the ( m = 10 ) maximum.
• the point p is between the ( m = 8 ) maximum and the ( m = 9 ) maximum.
• the point p is between the ( m = 16 ) maximum and the ( m = 17 ) maximum.

Explanation:

Step1: Recall double - slit formula

The formula for the position of the \(m\) - th maximum in a double - slit experiment is \(y = m\frac{\lambda L}{d}\), where \(y\) is the distance from the central maximum, \(\lambda\) is the wavelength of light, \(L\) is the distance from the slits to the screen, and \(d\) is the separation between the two slits. We can rearrange this formula to solve for \(m\): \(m=\frac{yd}{\lambda L}\)

Step2: Convert units

First, convert all units to SI units.

• \(y = 81.5\space cm=0.815\space m\)
• \(\lambda = 651\space nm = 651\times10^{-9}\space m\)
• \(d = 31.2\times 10^{-6}\space m\)
• \(L = 4.43\space m\)

Step3: Substitute values into the formula

Substitute the values of \(y\), \(d\), \(\lambda\), and \(L\) into the formula for \(m\):

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Since \(m\approx8.82\), the point \(P\) is between the \(m = 8\) maximum and the \(m=9\) maximum.

Answer:

The point \(P\) is between the \(m = 8\) maximum and the \(m = 9\) maximum.