In a Young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength lambda

Q: Why do we add 1 at the end of the calculation?

The formula $d \sin\theta = n\lambda$ gives you the number of fringes on one side of the central maximum. The $+1$ accounts for the central bright fringe itself, which is located at $n=0$.

Q: What if $n_{\text{max}}$ turned out to be a decimal, like 320.5?

Since the order of a fringe must be an integer, you would take the greatest integer less than or equal to that value. In the case of 320.5, the maximum complete fringe observed would be the 320th one.

Q: Why do we use $\sin\theta$ instead of $\tan\theta$ here?

The exact path difference is $d \sin\theta$. We only use the approximation $\tan\theta \approx \sin\theta$ for very small angles. Since $30^\circ$ is a large angle, we must use the exact formula.

Q: Are these fringes equally spaced on the screen?

No, they are not. The linear position on the screen is $y = D \tan\theta$. For large angles, $\tan\theta$ is not linear, so the fringes get further apart as you move away from the center.

Q: Does the distance to the screen ($D$) affect the number of fringes?

Not in this case! The question asks for the number of fringes within a specific *angular* range. The angular position $\theta$ depends only on $d$ and $\lambda$, making it completely independent of the screen distance $D$.

Question

In a Young's double slit experiment, the slits are placed $0.320 	ext{ mm}$ apart. Light of wavelength $\lambda = 500 	ext{ nm}$ is incident on the slits. The total number of bright fringes that are observed in the angular range $-30^\circ \le 	heta \le 30^\circ$ is

Options:

(A) 320

(B) 321

(C) 640

(D) 641

Sigma Percentile · Accepted Answer

The Young's Double Slit Experiment (YDSE) is a beautiful demonstration of the wave nature of light. When coherent light passes through two closely spaced slits, it creates an interference pattern of alternating bright and dark fringes on a screen. But how many fringes can we actually see within a specific viewing angle? Let's dive into the physics and find out!

### Analyzing the Setup

Imagine you are standing in front of a YDSE setup. The two slits are separated by a tiny distance, $d = 0.320 	ext{ mm}$. We are shining a monochromatic light of wavelength $\lambda = 500 	ext{ nm}$ onto these slits.

The question asks for the total number of bright fringes within an angular range of $-30^\circ \le 	heta \le 30^\circ$. This means we are looking at a specific "window" on the screen, spanning $30^\circ$ above and $30^\circ$ below the central axis.

### The Master Equation

To find the bright fringes (maxima), we need to recall the condition for constructive interference. The path difference between the light waves arriving from the two slits must be an integer multiple of the wavelength. Mathematically, this is expressed as:

$$d \sin	heta = n\lambda$$

Here, $n$ is the order of the fringe ($n = 0, 1, 2, \dots$). The central maximum corresponds to $n = 0$, the first bright fringe to $n = 1$, and so on.

### Finding the Maximum Order

To find the total number of fringes within our $30^\circ$ window, we first need to determine the highest order fringe, $n_{	ext{max}}$, that can form exactly at the boundary angle $	heta = 30^\circ$.

Rearranging our master equation to solve for $n$, we get:

$$n_{	ext{max}} = \frac{d \sin 30^\circ}{\lambda}$$

Now, let's carefully substitute our given values. Remember to convert everything into standard SI units (meters) to avoid any silly mistakes!

$$d = 0.320 	imes 10^{-3} 	ext{ m}$$
$$\lambda = 500 	imes 10^{-9} 	ext{ m}$$
$$\sin 30^\circ = 0.5$$

Plugging these in:

$$n_{	ext{max}} = \frac{(0.320 	imes 10^{-3}) 	imes 0.5}{500 	imes 10^{-9}}$$

$$n_{	ext{max}} = \frac{0.160 	imes 10^{-3}}{5 	imes 10^{-7}}$$

$$n_{	ext{max}} = \frac{160}{0.5} = 320$$

This tells us that exactly at the $30^\circ$ mark, the 320th bright fringe is formed.

### The Final Count

Now, we must be careful. $n_{	ext{max}} = 320$ is just the number of fringes on **one side** of the central maximum.

Because the interference pattern is symmetric, there will be another 320 fringes on the opposite side (in the $-30^\circ$ direction). And we must never forget the central maximum itself, which sits right in the middle at $n = 0$.

Therefore, the total number of bright fringes is:

$$	ext{Total Fringes} = 2n_{	ext{max}} + 1$$

$$	ext{Total Fringes} = 2(320) + 1 = 641$$

**The total number of bright fringes observed in the given angular range is 641.**

Animated Solution for Physics - Optics: 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

Select Answer:

Visualized Solution

YDSE Setup and Parameters

Condition for Bright Fringes $d sin θ = nλ$

Maximum Order $n_{max}$ at $θ = 3 0^{\circ}$

Calculating $n_{max}$

Total Number of Fringes

Conclusion

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

Analyzing the Setup

The Master Equation

Finding the Maximum Order

The Final Count

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