Question

in a double - slit diffraction experiment, two slits of width 11.1×10⁻⁶ m are separated by a distance of 32.4×10⁻⁶ m, and the wavelength of the incident light is 684 nm. the diffraction pattern is viewed on a screen 4.69 m from the slits. assume ( i_p ) is the intensity at a point p, a distance ( y = 68.9 ) 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 = 12 ) maximum and the ( m = 13 ) maximum.
• the point p is between the ( m = 6 ) maximum and the ( m = 7 ) maximum.
• the point p is between the ( m = 7 ) maximum and the ( m = 8 ) maximum.

Explanation:

Step1: Recall double - slit formula

The position of the \(m\) - th maximum in a double - slit interference pattern is given by \(y = m\frac{\lambda L}{d}\), where \(\lambda\) is the wavelength of light, \(L\) is the distance between the slits and the screen, \(d\) is the separation between the two slits, and \(m = 0,1,2,\cdots\)

First, convert all units to SI units:

• \(\lambda=684\space nm = 684\times10^{-9}\space m\)
• \(y = 68.9\space cm=0.689\space m\)
• \(L = 4.69\space m\)
• \(d = 32.4\times10^{-6}\space m\)

Step2: Solve for \(m\)

We can rearrange the formula \(y = m\frac{\lambda L}{d}\) to solve for \(m\):
\(m=\frac{y d}{\lambda L}\)

Substitute the values:
\(m=\frac{0.689\times32.4\times 10^{-6}}{684\times10^{-9}\times4.69}\)

First, calculate the numerator: \(0.689\times32.4\times 10^{-6}=0.689\times32.4\times10^{-6}=22.3236\times 10^{-6}\)

Then, calculate the denominator: \(684\times10^{-9}\times4.69 = 684\times4.69\times10^{-9}=3207.96\times 10^{-9}=3.20796\times 10^{-6}\)

Now, divide the numerator by the denominator:
\(m=\frac{22.3236\times 10^{-6}}{3.20796\times 10^{-6}}\approx6.96\)

Answer:

The point \(P\) is between the \(m = 6\) maximum and the \(m = 7\) maximum.