In a Young's double slit experiment, 16 fringes are observed in a certain segment of the screen when

Q: Why is the segment length considered constant?

Because the problem states we are observing the 'same segment of the screen'. We are looking at a fixed physical region, so its length $L$ does not change when we swap the light source.

Q: Why do $D$ and $d$ cancel out in the equation?

The physical setup of the slits and the screen hasn't been moved. Therefore, the distance between the slits ($d$) and the distance to the screen ($D$) remain identical for both wavelengths and cancel out mathematically.

Q: Does a smaller wavelength always mean more fringes in a given space?

Yes! A smaller wavelength results in a smaller fringe width ($\beta$). Since the fringes are thinner, you can pack more of them into the same fixed segment length.

Q: Do we need to convert nanometers to meters before calculating?

No, it's not necessary here. Because we are setting up a ratio ($N_1 \lambda_1 = N_2 \lambda_2$), the units of wavelength on both sides will cancel each other out, saving you calculation time.

Q: What happens to the central maximum when the wavelength changes?

The central maximum stays in the exact same position (path difference is zero for all wavelengths there). The fringes just compress or expand symmetrically around it.

Question

In a Young's double slit experiment, 16 fringes are observed in a certain segment of the screen when light of wavelength 700 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen would be

Options:

(A) 24

(B) 30

(C) 18

(D) 28

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a classic Young's Double Slit Experiment (YDSE) setup. We have a screen placed at a distance $D$ from the slits, and the slits themselves are separated by a distance $d$. Let's focus our attention on a specific segment of length $L$ on this screen.

Now, we know that the width of a single fringe, denoted by $\beta$, is given by the formula $\beta = \frac{\lambda D}{d}$. If there are $N$ fringes perfectly fitting into our chosen segment, then the total length $L$ of that segment is simply the number of fringes multiplied by the width of one fringe.

Mathematically, this is expressed as $L = N 	imes \beta$.

### The Master Equation

Since we are looking at the exact same physical segment on the screen, its length $L$ remains absolutely constant, even if we decide to change the light source. This is the crucial insight!

So, we can equate the expressions for the two different wavelengths:

$$L = N_1 \frac{\lambda_1 D}{d} = N_2 \frac{\lambda_2 D}{d}$$

Notice how the distance to the screen $D$ and the slit separation $d$ are the identical in both cases. Because the physical apparatus hasn't moved, they beautifully cancel out from both sides of our equation.

This leaves us with a very elegant and simple inverse relationship:

$$N_1 \lambda_1 = N_2 \lambda_2$$

### Final Calculation

Let's substitute the values given in the problem. Initially, we have $16$ fringes with a wavelength of $700 	ext{ nm}$. We need to find the new number of fringes, $N_2$, when the wavelength is changed to a shorter $400 	ext{ nm}$.

$$16 	imes 700 = N_2 	imes 400$$

Rearranging the equation to solve for $N_2$, we get:

$$N_2 = \frac{16 	imes 700}{400}$$

The zeros cancel out nicely. Sixteen divided by four is four, and four times seven gives us our final answer.

**$$N_2 = 28$$**

So, exactly $28$ fringes will be observed in the same segment. The correct option is (d).

Animated Solution for Physics - Optics: In a Young's double slit experiment, 16 fringes are observed in a certain segment of the screen when light of wavelength 700 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen would be

Select Answer:

Visualized Solution

\text{Young's Double Slit Experiment}

\text{Fringe Width and Segment Length}

\text{Equating Segment Lengths}

\text{Simplifying the Relation}

\text{Substituting the Values}

\text{Solving for } N_2

\text{Final Calculation}

\text{Food for Thought}

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

Analyzing the Setup

The Master Equation

Final Calculation

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