Question

In young's double slit experiment the phase difference is constant between two sources is \((\pi / 2)\). The intensity at a point equidistant from the slits in terms of max. intensity \(\mathrm{I}_{0}\) is (A) \(3 \mathrm{I}_{0}\) (B) \(\left(\mathrm{I}_{0} / 2\right)\) (C) I_{0 } (D) \(\left(3 \mathrm{I}_{0} / 4\right)\)

Short answer

In Young's double slit experiment with a phase difference of \(\frac{\pi}{2}\), the intensity at a point equidistant from the slits in terms of maximum intensity \(I_0\) is (C) \(I_0\).

Step-by-step solution

Recall the formula for the interference pattern intensity.

In Young's double slit experiment, the intensity of the interference pattern at a point P can be calculated using the formula: \(I_P = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\delta)\) Where; \(I_P\) = Intensity of the interference pattern at point P, \(I_1\) and \(I_2\) = Individual intensities of the two sources, \(\delta\) = Phase difference between the two sources. Since the two sources are identical, we can write: \(I_1 = I_2 = \frac{I_0}{2}\), where \(I_0\) is the maximum intensity.

Calculate the interference pattern intensity at the given point.

Now, we can substitute the values of the given phase difference and individual intensities into the formula for the interference pattern intensity: \(I_P = \frac{I_0}{2} + \frac{I_0}{2} + 2\sqrt{\frac{I_0}{2} \cdot \frac{I_0}{2}} \cos(\frac{\pi}{2})\) Since the cosine of \(\frac{\pi}{2}\) is equal to 0, the equation simplifies to: \(I_P = \frac{I_0}{2} + \frac{I_0}{2}\) Adding the two terms, we get: \(I_P = I_0\)

Match the calculated intensity to the correct answer choice.

The calculated intensity at the point equidistant from the slits is \(I_P = I_0\). This corresponds to the answer choice (C), which indicates that the intensity at the point equidistant from the slits in terms of maximum intensity is \(I_0\). Therefore, the correct answer is (C) \(I_0\).

Key concepts

Interference Pattern

When light waves pass through two narrow slits, they overlap and create a series of bright and dark spots on a screen. This pattern is called an interference pattern. It's a beautiful demonstration of the wave nature of light.

In Young's double slit experiment, the bright spots, or maxima, occur where the light waves from the two slits meet in phase and reinforce each other. The dark spots, or minima, arise where the waves are out of phase and cancel each other out.

To visualize this, imagine two sets of ripples in a pond meeting. Where the crests of the ripples align perfectly, the wave height is greater. This is similar to where the bright spots appear on the interference pattern.

• The alternating bright and dark bands assist in measuring wavelength.
• This experiment supports the wave theory of light.

Understanding interference patterns is crucial for applications like diffraction gratings, which are used to measure light wavelengths.

Phase Difference

Phase difference is the key to understanding how light waves interact in the double slit experiment. It refers to the difference in the phase angle of the waves when they meet at a point.

In simple terms, if you imagine one wave as starting a bit later than the other, the amount by which it lags behind is the phase difference. This determines whether the waves will add up to create a bright band or cancel out for a dark band.

In Young's experiment, a constant phase difference implies a fixed relationship in the timing of the wave crests. When the phase difference is \( \frac{\pi}{2} \), for example, the waves are a quarter of a cycle out of sync.

• Phase difference is scientifically measured in radians.
• Determines the position of interference maxima and minima.

This concept is vital for technologies using wave interference, such as noise-cancelling headphones.

Intensity Formula

The intensity of light at any point in an interference pattern can be calculated using a specific formula. This formula takes into account the phase difference and the intensity of light from each slit.

The formula is given by:\[ I_P = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\delta) \]where: - \( I_P \) is the intensity at a point, - \( I_1 \) and \( I_2 \) are the intensities from each slit, - \( \delta \) is the phase difference. The cosine term is crucial as it captures the effect of phase difference, showing how the two beams reinforce or cancel each other.When \( \delta = \frac{\pi}{2} \), like in the exercise, \( \cos(\frac{\pi}{2}) \) becomes 0, simplifying the equation to just the sum of individual intensities, showing how evenly light is distributed when out of phase.

• Helps in calculating locations of maximum and minimum intensity.
• Demonstrates the principle of superposition of waves.

This formula is essential in optical instruments for focusing and filtering light.