A light source, which emits two wavelengths lambda_1 = 400 nm and lambda_2 = 600 nm, is used in a Yo

Q: Why is the number of fringes inversely proportional to the fringe width?

Think of it like fitting boxes on a shelf. If the boxes (fringe widths) are wider, fewer of them will fit in the same shelf space (distance $y$).

Q: How do we know when a maximum and a minimum overlap?

They overlap when their distance $y$ from the central maximum is exactly the same. By equating their position formulas, we can check if the condition holds true.

Q: What is the difference between linear separation and angular separation?

Linear separation is the actual distance $y$ on the screen, while angular separation $\theta$ is the angle subtended by the fringe at the slits. Angular separation is simply $\frac{\lambda}{d}$ and doesn't depend on the screen distance $D$.

Q: Why do we use $(n - 1/2)$ for minima instead of $(n + 1/2)$?

It depends on the convention. If we start counting from $n=1$ for the first minimum, the path difference is $\frac{\lambda}{2}$, which matches $\left(1 - \frac{1}{2}\right)\lambda$. If we used $\left(n+\frac{1}{2}\right)$, the first minimum would correspond to $n=0$.

Q: Can we see this overlap with the naked eye?

Yes! If you perform this experiment with a mix of 400 nm (violet) and 600 nm (orange-red) light, you will see a distinct color pattern where the bright orange-red fringe perfectly fills the dark gap of the violet fringe.

Question

A light source, which emits two wavelengths $\lambda_1 = 400	ext{ nm}$ and $\lambda_2 = 600	ext{ nm}$, is used in a Young's double-slit experiment. If recorded fringe widths for $\lambda_1$ and $\lambda_2$ are $\beta_1$ and $\beta_2$ and the number of fringes for them within a distance $y$ on one side of the central maximum are $m_1$ and $m_2$, respectively, then

Options:

(A) $\beta_2 > \beta_1$

(B) $m_1 > m_2$

(C) from the central maximum, 3rd maximum of $\lambda_2$ overlaps with 5th minimum of $\lambda_1$

(D) the angular separation of fringes of $\lambda_1$ is greater than $\lambda_2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Welcome to this beautiful problem on Young's Double Slit Experiment (YDSE). We are given a light source that emits two distinct wavelengths simultaneously: $\lambda_1 = 400 ext{ nm}$ and $\lambda_2 = 600 ext{ nm}$. Our goal is to analyze the interference pattern they create on the screen and evaluate the four given options. ### Fringe Widths and Counts First, let's look at the fringe width, denoted by $\beta$. The formula for fringe width is: $$\beta = \frac{\lambda D}{d}$$ Since the fringe width is directly proportional to the wavelength, and we know that $\lambda_2 > \lambda_1$ ($600 ext{ nm} > 400 ext{ nm}$), it is obvious that $\beta_2 > \beta_1$. Therefore, **Option (a) is absolutely correct**. Now, suppose we take a fixed distance $y$ on the screen. How many fringes will fit in this distance? The number of fringes $m$ is simply the total distance $y$ divided by the fringe width $\beta$: $$m = \frac{y}{\beta}$$ Because $\beta_2$ is larger, its fringes are wider, meaning fewer of them will fit in the same distance $y$. Thus, $m_1 > m_2$. **Option (b) is also correct**. ### The Overlap Condition Let's check the overlapping condition given in option (c). Where is the 3rd maximum of $\lambda_2$ located? Using the formula for the position of maxima, $y = n \frac{\lambda D}{d}$, we plug in $n = 3$ and $\lambda_2 = 600 ext{ nm}$: $$y_2 = 3 \frac{(600) D}{d} = \frac{1800 D}{d}$$ What about the 5th minimum of $\lambda_1$? The formula for the position of minima is $y = \left(n - \frac{1}{2} ight) \frac{\lambda D}{d}$. For the 5th minimum, $n = 5$, so we get $4.5$ times $\frac{\lambda_1 D}{d}$. Substituting $400 ext{ nm}$ for $\lambda_1$: $$y_1 = 4.5 \frac{(400) D}{d} = \frac{1800 D}{d}$$ Look closely at the results. Both positions are exactly the same! This means the 3rd maximum of $\lambda_2$ perfectly overlaps with the 5th minimum of $\lambda_1$. So, **Option (c) is indeed correct**. Isn't it beautiful how the math aligns perfectly? ### Angular Separation Finally, let's evaluate the angular separation of the fringes, which is given by: $$ heta = \frac{\lambda}{d}$$ Since $\lambda_1 < \lambda_2$, the angular separation for $\lambda_1$ must be smaller than that for $\lambda_2$ (i.e., $ heta_1 < heta_2$). Option (d) claims the opposite, so it is incorrect. Our final correct options are **(a), (b), and (c)**.

Select Answer:

Visualized Solution

\text{Young's Double Slit Setup}

\text{Fringe Width Comparison}

\text{Number of Fringes}

\text{Position of 3rd Maximum of } \lambda_2

\text{Position of 5th Minimum of } \lambda_1

\text{Overlapping Condition}

\text{Angular Separation}

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

Analyzing the Setup

Fringe Widths and Counts

The Overlap Condition

Angular Separation

Similar Questions

In a Young's double slit experiment, the separation between the two slits is $d$ and the wavelength of the light is $λ$ . The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice (s).

In a Young's double slit experiment, two slits are separated by $2 mm$ and the screen is placed one metre away. When a light of wavelength $500 nm$ is used, the fringe separation will be

In a Young's double slit experiment, the slits are placed $0.320 mm$ apart. Light of wavelength $λ = 500 nm$ is incident on the slits. The total number of bright fringes that are observed in the angular range $- 3 0^{\circ} \leq θ \leq 3 0^{\circ}$ is

In a Young's double slit experiment, the separation between the slits is $0.15 mm$ . In the experiment, a source of light of wavelength $589 nm$ is used and the interference pattern is observed on a screen kept $1.5 m$ away. The separation between the successive bright fringes on the screen is