In a Young's double slit experiment, 12 fringes are observed to be formed in a certain segment of th

Q: Why does the number of fringes increase when the wavelength decreases?

Fringe width is directly proportional to the wavelength. A smaller wavelength produces narrower fringes, allowing more of them to fit into the same fixed segment on the screen.

Q: Can we use this logic if the screen distance $D$ is also changed?

No. If $D$ changes, the fringe width changes differently. You would need to equate $y = n_1 \frac{\lambda_1 D_1}{d} = n_2 \frac{\lambda_2 D_2}{d}$.

Q: What happens if the entire setup is immersed in water?

The wavelength of light decreases by a factor of the refractive index ($\mu$). This reduces the fringe width, so the number of fringes in the same segment will increase by a factor of $\mu$.

Q: Is the central maximum included in this count?

The problem states 'in a certain segment'. It doesn't matter if the central maximum is included or not, as long as the segment length $y$ is a multiple of the fringe width.

Q: Why do we equate $n_1 \lambda_1 = n_2 \lambda_2$?

Because the segment length $y = n \beta = n \frac{\lambda D}{d}$ is constant. Since $D$ and $d$ are fixed, $n \lambda$ must be constant.

Question

In a Young's double slit experiment, $12$ fringes are observed to be formed in a certain segment of the screen when light of wavelength $600	ext{ nm}$ is used. If the wavelength of light is changed to $400	ext{ nm}$, number of fringes observed in the same segment of the screen is given by

Options:

(A) 12

(B) 18

(C) 24

(D) 30

Sigma Percentile · Accepted Answer

### The Fixed Window on the Screen

Imagine you are looking at a Young's Double Slit Experiment setup. Instead of looking at the entire screen, you place a rectangular frame over a specific portion of it. We will call the length of this fixed window **$y$**.

Whatever happens to the light source, we are only interested in counting the number of fringes that fit perfectly inside this window.

### The First Scenario: Red Light

Initially, we use a light source with a wavelength of **$\lambda_1 = 600 	ext{ nm}$**. When we look through our window of length $y$, we count exactly **$n_1 = 12$** fringes.

How do we relate the length of the window to the fringes? It is simple geometry. The total length $y$ is just the number of fringes multiplied by the width of a single fringe ($\beta$).

$$y = n_1 \beta_1$$

We know from the theory of interference that the fringe width is given by $\beta = \frac{\lambda D}{d}$, where $D$ is the distance to the screen and $d$ is the slit separation. Substituting this, we get:

$$y = n_1 \frac{\lambda_1 D}{d}$$

### The Master Equation

Now, we change the light source to a shorter wavelength, **$\lambda_2 = 400 	ext{ nm}$**.

Here is the crucial insight: **the window length $y$, the screen distance $D$, and the slit separation $d$ have not changed.** They are all constants.

If we write the equation for the second scenario, it looks like this:

$$y = n_2 \frac{\lambda_2 D}{d}$$

Since the left-hand side ($y$) is the same for both scenarios, we can equate the right-hand sides:

$$n_1 \frac{\lambda_1 D}{d} = n_2 \frac{\lambda_2 D}{d}$$

The constants $D$ and $d$ cancel out beautifully, leaving us with a very elegant and powerful relationship:

$$n_1 \lambda_1 = n_2 \lambda_2$$

This equation tells us that the number of fringes is inversely proportional to the wavelength.

### Final Calculation

Let's substitute the values we know into our master equation:

$$12 	imes 600 = n_2 	imes 400$$

Solving for the new number of fringes, $n_2$:

$$n_2 = \frac{12 	imes 600}{400}$$

$$n_2 = 18$$

So, exactly **18 fringes** will now fit into the same segment on the screen.

Think about the physical intuition here. A shorter wavelength (400 nm compared to 600 nm) produces narrower fringes. If the fringes are thinner, you can naturally pack more of them into the exact same space!

Select Answer:

Visualized Solution

$y$

$λ_{1} = 600 nm$

$y = n β$

$β = \frac{λ D}{d}$

$n_{1} λ_{1} = n_{2} λ_{2}$

$12 \times 600 = n_{2} \times 400$

$n_{2} = 18$

$Conclusion$

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

The Fixed Window on the Screen

The First Scenario

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ1​=600 nm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=nβ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β=dλD​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n1​λ1​=n2​λ2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 12×600=n2​×400

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n2​=18

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion