Question

Coherent monochromatic light with wavelength \(\lambda=514 \mathrm{nm}\) is incident on two slits that are separated by a distance \(d=0.500 \mathrm{~mm} .\) The intensity of the radiation at a screen \(2.50 \mathrm{~m}\) away from each slit is \(180.0 \mathrm{~W} / \mathrm{cm}^{2} .\) Deter-

Short answer

Answer: The intensity of light at an angle of \(30^{\circ}\) from the central maximum is \(0 \, \mathrm{W/cm}^2\).

Step-by-step solution

Calculate the path difference at \(30^{\circ}\)

To find the path difference, we can use the formula \(d \, \sin{\theta}\), where \(d\) is the distance between the slits and \(\theta\) is the given angle. Here, \(d=0.500 \mathrm{~mm}\) and \(\theta = 30^{\circ}\). Calculate the path difference: $$ \Delta{x} = d \, \sin{\theta} = 0.500 \times 10^{-3}\, \mathrm{m} \times \sin{30^{\circ}} = 0.250 \times 10^{-3}\, \mathrm{m}. $$

Calculate the phase difference

To calculate the phase difference, we can use the formula \(\delta = \dfrac{2\pi}{\lambda} \Delta{x}\), where \(\lambda = 514 \,\mathrm{nm}\). Calculate the phase difference: $$ \delta = \dfrac{2\pi}{514 \times 10^{-9}\, \mathrm{m}} \times 0.250 \times 10^{-3}\, \mathrm{m} = \dfrac{2\pi}{514 \times 10^{-9}\, \mathrm{m}} \times 250 \times 10^{-9}\, \mathrm{m} = \pi. $$

Calculate the resultant amplitude

To obtain the resultant amplitude, we can apply the phase difference to the amplitude formula: $$ A_r = 2A\cos{\dfrac{\delta}{2}} = 2A\cos{\dfrac{\pi}{2}} = 2A \times 0 = 0, $$ where \(A_r\) is the resultant amplitude and \(A\) is the amplitude of each wave from the two slits. The resultant amplitude turns out to be zero.

Calculate the intensity at \(30^{\circ}\) angle

Finally, we can find the intensity using the formula \(I = kA_r^2\), where \(k\) is a proportionality constant. Since the resultant amplitude is zero, the intensity will also be zero: $$ I = k \times 0^2 = 0. $$ At an angle of \(30^{\circ}\) from the central maximum, the intensity is \(0 \, \mathrm{W/cm}^2\).

Key concepts

Path Difference

In understanding wave interference, path difference is a key concept. It involves measuring how much longer one wave travels compared to another. Simply put, it's the extra distance one wave has to cover to get from a starting point to a shared endpoint. We calculate this difference using the formula: \( \Delta x = d \sin{\theta} \), where \( d \) is the gap between two sources (like slits) and \( \theta \) is the angle of interest.

• In the given exercise, the path difference was calculated using \( d = 0.500 \text{ mm} \) and \( \theta = 30^{\circ} \).
• This resulted in a path difference of \( 0.250 \times 10^{-3} \text{ m} \).

This difference in path length is crucial because it determines how waves interfere with each other—constructively or destructively.
Without understanding path difference, predicting interference patterns would be nearly impossible.

Phase Difference

Phase difference describes how "out of step" two waves are. It's a measure of the offset, expressed in radians, between two wave's peaks and troughs. To calculate phase difference, we use the formula: \( \delta = \dfrac{2\pi}{\lambda} \Delta x \). Here, \( \lambda \) is the wavelength of the light, and \( \Delta x \) is the path difference we just calculated.

• Using a wavelength of \( 514 \text{ nm} \) and the path difference already found, the phase difference was \( \pi \) radians.
• This corresponds to the waves being exactly half a wavelength out of phase, leading to destructive interference.

Understanding phase difference helps predict whether waves will enhance each other or cancel out. At \( \delta = \pi \), complete cancellation occurs.

Amplitude Calculation

Amplitude is a measure of the wave's strength or intensity. When two waves interfere, their resultant amplitude can be calculated by considering their phase difference. The formula used here is: \( A_r = 2A\cos{\dfrac{\delta}{2}} \), where \( A \) is the initial amplitude of each wave, and \( \delta \) is the phase difference.

• In the case of the exercise, with \( \delta = \pi \), the \( \cos{\dfrac{\pi}{2}} \) equals zero, resulting in a resultant amplitude of zero.
• This zero amplitude indicates complete destructive interference, leading to a resulting wave of zero intensity.

Calculating amplitude is crucial for predicting the power and impact of wave interference, helping students visualize concepts like loud and quiet spots in sound or light waves.