Question

Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. (The phase difference between beam is \((\pi / 2)\) at the point \(\mathrm{A}\) and \(\pi\) at point \(\mathrm{B}\).) (A) I (B) \(4 \mathrm{I}\) (C) \(2 \mathrm{I}\) (D) \(6 \mathrm{I}\)

Short answer

The intensities at points A and B due to the resulting amplitudes are 5I and I, respectively. None of the given options match these results, so there might be an error in the question or it is incomplete.

Step-by-step solution

Calculate the amplitudes of the beams

To start, we need to find the amplitudes of the two beams. The intensity I of a wave is proportional to the square of its amplitude A. Thus, we can deduce that \(I=kA^2\) and \(4I=kA'^2\), where A and A' are the amplitudes of the two beams, and k is a constant. Now, we can calculate the amplitude A' of the second beam: \(A'^{2} = \frac{4I}{k} = 4A^{2}\) So, \(A'=2A\).

Determine the intensities at points A and B

Now, we'll determine the intensities at points A and B. To do this, we'll use the resulting amplitude for the two beams at each point and their given phase differences: At point A, the phase difference is \((\pi / 2)\). The resulting amplitude at point A can be calculated using the following formula: \(R_A = \sqrt{A^2 + A'^2 + 2AA'\cos(\pi/2)}\) At point B, the phase difference is \(\pi\). The resulting amplitude at point B can be calculated using the same formula: \(R_B = \sqrt{A^2 + A'^2 + 2AA'\cos(\pi)}\)

Compute the resulting amplitudes

Now, substitute the values of A and A' into the formulas for the resulting amplitudes at A and B: At point A: \(R_A = \sqrt{A^2 + (2A)^2 + 2A(2A)\cos(\pi/2)}\) \(R_A = \sqrt{A^2 + 4A^2 + 0}\) \(R_A = \sqrt{5A^2}\) \(R_A = A\sqrt{5}\) At point B: \(R_B = \sqrt{A^2 + (2A)^2 + 2A(2A)\cos(\pi)}\) \(R_B = \sqrt{A^2 + 4A^2 - 4A^2}\) \(R_B = \sqrt{A^2}\) \(R_B = A\)

Calculate the resulting intensities

Finally, find the intensities at points A and B due to the resulting amplitudes R_A and R_B: Intensity at point A: \(I_A = kR_A^2 = k(5A^2) = 5I\) Intensity at point B: \(I_B = kR_B^2 = k(A^2) = I\) So, the intensities at points A and B are 5I and I, respectively. Looking at the given options, none of them match these results, so there might be an error in the question, or it is incomplete.

Key concepts

Intensity of Light

The intensity of light is an important concept in understanding optical interference. It essentially measures the amount of energy that passes through a certain area in a set period of time. In simpler terms, it's like describing how "bright" the light appears to our eyes. The intensity, denoted as \(I\), is directly related to the amplitude of the wave, where the relationship is quadratic.

• This means if the amplitude of a light wave doubles, its intensity doesn't just double—it increases by a factor of four.


• Mathematically, this is expressed as \(I \propto A^2\).

Understanding this helps explain why, in interference patterns, the brightness of the fringes can vary greatly. When two beams of light interfere with each other, their intensities combine in a way that can either amplify or diminish the resulting light due to their phase differences.

Amplitude of Waves

The amplitude of a wave is a crucial factor that impacts the light's intensity. It's akin to the "height" of the wave, though in light, it refers to the extent of oscillation of the electric and magnetic fields. The larger the amplitude, the more energy the wave carries, which translates directly to a greater intensity.

In situations involving interference, such as the problem we are examining, the amplitudes of the individual beams interact with each other.

• For two light beams with intensities \(I\) and \(4I\), the amplitudes are related by \(A\) and \(2A\), respectively, as \(A' = 2A\) for the beam with higher intensity.

This relationship highlights that changes in amplitude directly affect the overall light pattern observed in interference experiments, leading to areas of constructive or destructive interference based on how these waves align.

Phase Difference

Phase difference is a fundamental concept when exploring optical interference. It refers to the difference in the phase angle between the peaks (or troughs) of two waves at a given point.

When two waves meet, they can interfere constructively or destructively depending on their phase difference:

• At a phase difference of \(0\) or multiples of \(2\pi\), waves align perfectly, leading to constructive interference and a bright fringe.


• Conversely, a phase difference of \(\pi\) causes perfect destructive interference, cancelling each other out and resulting in a dark fringe.

In the given exercise, at point A, with a phase difference of \(\pi/2\), the light waves partially interfere constructively, leading to intensity changes that do not reach the maximum possible constructive interference. Meanwhile, at point B, with a phase difference of \(\pi\), the interference is destructive, perfectly cancelling out any additional effect on the intensity beyond the dominant wave's contribution.