What are the conditions for interference of light?

The primary conditions are that the sources must be coherent, monochromatic (of the same frequency/wavelength), and have a constant phase difference.

Why do red and green light not interfere?

Red and green light have different frequencies. Interference requires sources of the same frequency to maintain a stationary pattern.

What is a coherent source?

Two sources are coherent if they emit light waves of the same frequency and have a zero or constant phase difference.

What happens if the phase difference changes rapidly?

If the phase difference changes randomly and rapidly, the interference pattern shifts so quickly that the eye only perceives a uniform average intensity.

What is the role of a color filter in YDSE?

A filter only allows specific wavelengths to pass. A red filter allows red light, and a green filter allows green light.

Is there any light on the screen in this case?

Yes, there is a general illumination resulting from the sum of intensities of red and green light, but no fringe pattern.

What is the result of this specific experiment?

The result is that there shall be no interference fringes.

Can white light produce interference fringes?

Yes, white light can produce a colored fringe pattern with a central white fringe if both slits allow the full spectrum.

Does the intensity change across the screen?

The intensity will be the sum of individual intensities ($I = I_1 + I_2$) without the interference term ($2\sqrt{I_1 I_2} \cos \phi$).

Why is the central fringe white in normal YDSE with white light?

Because at the center, the path difference is zero for all wavelengths, leading to constructive interference for all colors.

In a Young's double slit experiment, the source is white light. One of the slits is covered by red filter and another by a green filter. In this case:

In a Young’s double slit experiment, the source is white light. One of the slits is covered by red filter and another by a green filter. In this case: | JEE Main Physics

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PhysicsWave Optics

QMCQ

In a Young’s double slit experiment, the source is white light. One of the slits is covered by red filter and another by a green filter. In this case:

A) there shall be alternate interference fringes of red and green.
B) there shall be an interference pattern for red distinct from that for green.
C) there shall be an interference pattern, where each fringe’s pattern center is green and outer edges is red.
D) there shall be no interference fringes.

✅ Correct Answer

Option D

✓

Solution Steps

Understand the Role of Filters

A red filter only allows light of the red wavelength ($\lambda_R$) to pass through, and a green filter only allows green wavelength ($\lambda_G$) light. After passing through the filters, the two slits act as two independent sources of different colors.

Condition for Interference (Frequency)

A fundamental condition for a stable interference pattern is that the two light sources must have the same frequency (or wavelength). Red light and green light have significantly different frequencies:

$$f_{red} \neq f_{green}$$

Phase Difference Analysis

For interference to occur, the phase difference $\Delta \phi$ between the two waves at any point on the screen must be constant over time. Since the frequencies are different, the phase difference changes continuously and very rapidly with time:

$$\Delta \phi(t) = (2\pi f_1 t + \phi_1) – (2\pi f_2 t + \phi_2)$$

Intensity Distribution

The intensity at any point is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)$. Because $\cos(\Delta \phi)$ oscillates so rapidly between $+1$ and $-1$, its average value over a very short time interval is zero.

Conclusion

Due to the lack of coherence and the difference in frequencies, no stationary interference pattern is formed. The screen will show uniform illumination (a mixture of red and green light) but no fringes.

Final Result: There shall be no interference fringes.

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Theory

1. Conditions for Sustained Interference

To observe a stable interference pattern, the following conditions must be met: (i) The sources must be coherent, meaning they maintain a constant phase difference. (ii) The sources must be monochromatic (emit light of a single wavelength). (iii) The amplitudes of the waves should ideally be equal for maximum contrast. In this problem, by using a red filter on one slit and a green filter on the other, we violate the requirement of monochromaticity between the two slits. Since red and green have different frequencies, they are non-coherent sources. Consequently, the interference term in the intensity equation averages to zero, leading to a loss of the fringe pattern.

2. Coherence and Frequency

Coherence is the property of waves that enables stationary interference. Two waves are coherent if they have the same frequency and a constant phase relationship. When waves of different frequencies ($f_1$ and $f_2$) overlap, the phase difference at any point in space becomes a function of time: $\Delta \phi = 2\pi(f_1 – f_2)t$. Because the frequency of light is extremely high ($\sim 10^{14}$ Hz), even a small difference in frequency causes the phase difference to cycle through $2\pi$ billions of times per second. No detector or human eye can resolve these rapid changes, resulting in a perceived uniform average intensity.

3. Intensity in Non-Coherent Superposition

When two coherent waves interfere, the resultant intensity is $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$. The third term is the “interference term,” which creates the bright and dark fringes. However, for non-coherent sources (like red and green light), the time average of $\cos \phi$ is zero. Therefore, the resultant intensity is simply the sum of the individual intensities: $I = I_1 + I_2$. This is why the screen appears uniformly lit. This principle explains why two independent light bulbs in a room do not produce an interference pattern on the walls; they are not coherent.

4. Color Filters in Wave Optics

A color filter is an optical device that selectively transmits light of a particular range of wavelengths while absorbing others. In YDSE, if both slits are covered by the same color filter (e.g., both red), we would observe a standard interference pattern consisting of red and dark fringes. The fringe width $\beta = \lambda D / d$ would be specific to that color. However, applying different filters effectively creates two completely different “channels” of light. Since light of one color does not have a stable phase relationship with light of another color, the “cross-talk” (interference) between them is non-existent in a stationary sense.

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FAQs

Will there be any light on the screen?

Yes, the screen will be illuminated with a uniform mixture of red and green light, but no dark or bright bands (fringes) will be visible.

What if both slits had red filters?

In that case, you would see a standard red interference pattern with a specific fringe width $\beta_R$.

Why is the frequency difference important?

Different frequencies mean the waves “slip” past each other in phase so fast that the interference pattern moves faster than we can see.

Can we ever get interference with two different colors?

Only as a transient “beat” phenomenon if the frequencies are very close, but not as a stationary YDSE pattern.

What does monochromatic mean?

Monochromatic light consists of a single wavelength or color.

What happens with white light and no filters?

You get a central white fringe surrounded by a few colored fringes, and then a uniform white background.

Is energy conserved if fringes disappear?

Yes, energy is still conserved. The total energy on the screen is the same; it’s just distributed uniformly instead of in bands.

Do red and green light waves even “meet”?

Yes, they physically overlap on the screen, but they don’t produce a stable phase-dependent reinforcement or cancellation.

Which has a larger fringe width, red or green?

Since $\beta \propto \lambda$ and $\lambda_{red} > \lambda_{green}$, red fringes are wider than green fringes (if they were formed).

Why is coherence necessary?

Coherence ensures the “peaks” and “troughs” of the two waves stay lined up at the same physical locations on the screen.

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