White light is used to illuminate the two slits in a Young's double slit experiment. The separation

Q: Why do missing wavelengths correspond to destructive interference?

When destructive interference occurs for a specific wavelength at a point, the intensity of light for that wavelength becomes zero. Hence, that color or wavelength is "missing" from the spectrum observed at that point.

Q: Why is the distance $y$ equal to $b/2$?

The central axis of the setup passes exactly midway between the two slits. Since the total distance between the slits is $b$, the distance from the central axis to either slit is exactly half of that, which is $b/2$.

Q: Can we use the exact path difference formula instead of the approximation?

Yes, the exact path difference is $\sqrt{d^2 + b^2} - d$. Using the binomial expansion for $d \gg b$, this simplifies to $d(1 + \frac{b^2}{2d^2}) - d = \frac{b^2}{2d}$, which perfectly matches our approximation $\frac{yb}{d}$ with $y = b/2$.

Q: What would happen if the point was at the center of the screen?

At the center of the screen ($y = 0$), the path difference is zero for all wavelengths. This means constructive interference occurs for all wavelengths, resulting in a bright white central fringe. No wavelengths would be missing there.

Q: Why do we use $(2n-1)\frac{\lambda}{2}$ instead of $(2n+1)\frac{\lambda}{2}$?

Both expressions represent odd multiples of $\frac{\lambda}{2}$ and are physically equivalent. Using $(2n-1)$ with $n=1, 2, 3 \dots$ naturally gives the first, second, and third minima, which is a standard convention.

Question

White light is used to illuminate the two slits in a Young's double slit experiment. The separation between the slits is $b$ and the screen is at a distance $d$ ($>> b$) from the slits. At a point on the screen directly in front of one of the slits, certain wavelengths are missing. Some of these missing wavelengths are

Options:

(A) $\lambda = \frac{b^2}{d}$

(B) $\lambda = \frac{2b^2}{d}$

(C) $\lambda = \frac{b^2}{3d}$

(D) $\lambda = \frac{2b^2}{3d}$

Sigma Percentile · Accepted Answer

The phenomenon of interference is one of the most beautiful demonstrations of the wave nature of light. In Young's Double Slit Experiment (YDSE), when white light—which is a continuous spectrum of all visible wavelengths—is passed through the two slits, it creates a stunning overlapping pattern of colored fringes on the screen.

However, at any specific point on the screen, not all colors will be bright. Some wavelengths will undergo perfectly destructive interference, meaning their intensity drops to zero. These are what we call the **missing wavelengths**. Let's dive into the mathematics of finding exactly which colors vanish at a specific point!

### Analyzing the Setup

Imagine you are looking at the YDSE apparatus. We have two slits, $S_1$ and $S_2$, separated by a tiny distance $b$. Far away, at a distance $d$ (where $d \gg b$), lies our screen.

The problem asks us to focus on a very specific point $P$ on the screen: the point directly in front of the upper slit $S_1$.

Because the central axis $O$ passes exactly midway between the two slits, the vertical distance from the central axis to the slit $S_1$ is exactly half of the slit separation. Therefore, the vertical position of our point $P$ is:
$$y = \frac{b}{2}$$

### The Master Equation for Path Difference

When light from $S_1$ and $S_2$ travels to point $P$, the ray from $S_2$ has to travel a slightly longer distance than the ray from $S_1$. This extra distance is the **path difference**, denoted by $\Delta x$.

For a screen placed very far away ($d \gg b$), the path difference at a vertical height $y$ is given by the standard approximation:
$$\Delta x = \frac{y \cdot b}{d}$$

Let's substitute the position of our point $P$ ($y = \frac{b}{2}$) into this formula:
$$\Delta x = \frac{\left(\frac{b}{2}\right) \cdot b}{d} = \frac{b^2}{2d}$$

This is the geometric path difference at point $P$.

### Condition for Missing Wavelengths

What does it mean for a wavelength to be "missing"? It means that the light waves of that specific wavelength arriving from $S_1$ and $S_2$ are exactly out of phase. The crest of one wave meets the trough of the other, canceling each other out completely.

This is the condition for **destructive interference** (a minimum). For destructive interference to occur, the path difference must be an odd multiple of half the wavelength:
$$\Delta x = (2n - 1)\frac{\lambda}{2}$$
where $n = 1, 2, 3, \dots$

### Final Calculation

Now, we simply equate our geometric path difference with the condition for destructive interference:
$$\frac{b^2}{2d} = (2n - 1)\frac{\lambda}{2}$$

Notice how the $2$ in the denominators on both sides beautifully cancels out! Let's rearrange the equation to solve for the wavelength $\lambda$:
$$\lambda = \frac{b^2}{(2n - 1)d}$$

This is our master formula for all the missing wavelengths at point $P$. To find the specific wavelengths, we just plug in integer values for $n$:

For $n = 1$ (the first missing wavelength):
$$\lambda_1 = \frac{b^2}{(2(1) - 1)d} = \frac{b^2}{d}$$

For $n = 2$ (the second missing wavelength):
$$\lambda_2 = \frac{b^2}{(2(2) - 1)d} = \frac{b^2}{3d}$$

For $n = 3$ (the third missing wavelength):
$$\lambda_3 = \frac{b^2}{(2(3) - 1)d} = \frac{b^2}{5d}$$

And the pattern continues. If we look at the given options, we can clearly see that $\lambda = \frac{b^2}{d}$ and $\lambda = \frac{b^2}{3d}$ are among the missing wavelengths.

Isn't it fascinating how simple geometry and wave mechanics perfectly predict which colors will vanish from the spectrum?

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Condition for Missing Wavelengths$

$Path Difference at Point P$

$Substituting y = \frac{b}{2}$

$Equating Path Differences$

$Finding the Missing Wavelengths$

$The Way Forward$

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

Analyzing the Setup

The Master Equation for Path Difference

Condition for Missing Wavelengths

Final Calculation

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Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Path Difference at Point P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting y=2b​

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

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

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