Young's double slit experiment is carried out by using green, red and blue light, one colour at one

Q: Why does the fringe width depend on the wavelength?

The fringe width is the distance between two consecutive bright or dark fringes. It is determined by the path difference between the light waves from the two slits. Since the path difference required for constructive or destructive interference is a multiple of the wavelength, a longer wavelength requires a larger physical distance on the screen to achieve the same phase difference, resulting in a wider fringe.

Q: Does the fringe width depend on the intensity of the light?

No, the fringe width $\beta = \frac{\lambda D}{d}$ only depends on the wavelength $\lambda$, the distance to the screen $D$, and the slit separation $d$. The intensity of the light affects the brightness of the fringes, but not their width or spacing.

Q: What would happen if we used white light instead of monochromatic light?

If white light is used, each wavelength will form its own interference pattern. The central maximum will be white because all wavelengths interfere constructively there. However, the other fringes will be colored because the fringe width varies with wavelength, causing the colors to overlap and separate, with violet closest to the center and red farthest away.

Q: Is the fringe width the same for all fringes in the pattern?

Yes, in a standard Young's Double Slit Experiment, the fringe width is constant for all fringes, provided the angle $\theta$ is small (which is typically the case when $D \gg d$).

Q: How can we remember the order of wavelengths for visible light?

You can use the acronym VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red). The wavelength increases as you move from Violet to Red. So, Violet has the shortest wavelength, and Red has the longest.

Question

Young's double slit experiment is carried out by using green, red and blue light, one colour at one time. The fringe widths recorded are $\beta_G, \beta_R$ and $\beta_B$ respectively.
Then,

Options:

(A) $\beta_G > \beta_B > \beta_R$

(B) $\beta_B > \beta_G > \beta_R$

(C) $\beta_R > \beta_B > \beta_G$

(D) $\beta_R > \beta_G > \beta_B$

Sigma Percentile · Accepted Answer

### The Magic of Interference

Imagine you are standing in a dark room, and you shine a light through two tiny, closely spaced slits. Instead of seeing just two bright lines on the wall, you see a beautiful pattern of alternating bright and dark bands. This is the magic of Young's Double Slit Experiment! These bands are called fringes, and their width—the distance between two consecutive bright or dark bands—is a key feature of the pattern.

### The Master Equation

The width of these fringes isn't random. It is governed by a very elegant equation:
$$ \beta = \frac{\lambda D}{d} $$
Here, $\beta$ is the fringe width, $\lambda$ is the wavelength of the light, $D$ is the distance from the slits to the screen, and $d$ is the distance between the two slits.

In our problem, the experimental setup remains exactly the same for all three colors of light. This means that $D$ and $d$ are constant. Therefore, the fringe width $\beta$ is directly proportional to the wavelength $\lambda$:
$$ \beta \propto \lambda $$

### The Colors of the Rainbow

Now, let's think about the colors of light. You might remember the acronym **VIBGYOR**, which stands for Violet, Indigo, Blue, Green, Yellow, Orange, and Red. This sequence represents the visible light spectrum arranged in order of increasing wavelength.

As we move from blue to green to red, the wavelength gets longer. So, we can write the relationship between their wavelengths as:
$$ \lambda_R > \lambda_G > \lambda_B $$

### The Final Conclusion

Since the fringe width is directly proportional to the wavelength, the color with the longest wavelength will produce the widest fringes, and the color with the shortest wavelength will produce the narrowest fringes.

Because red light has the longest wavelength among the three, its fringe width $\beta_R$ will be the largest. Green light comes next, so its fringe width $\beta_G$ will be in the middle. Finally, blue light has the shortest wavelength, so its fringe width $\beta_B$ will be the smallest.

Putting it all together, we get the final relationship:
$$ \beta_R > \beta_G > \beta_B $$

This perfectly matches option (d). It's fascinating how a simple property like wavelength can directly dictate the physical size of the interference pattern we see on the screen!

Animated Solution for Physics - Optics: Young's double slit experiment is carried out by using green, red and blue light, one colour at one time. The fringe widths recorded are $β_{G}, β_{R}$ and $β_{B}$ respectively. Then,

Select Answer:

Visualized Solution

$Fringe Width Formula$

$Wavelengths of Colors$

$Comparing Fringe Widths$

The Sigma Insight: Interference and Young's Double-Slit Experiment

The Magic of Interference

The Master Equation

The Colors of the Rainbow

The Final Conclusion

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