A Young's double slit experiment is performed using monochromatic light of wavelength lambda. The in

Q: Why did we assume the intensity of both slits is the same ($I_0$)?

In a standard Young's Double Slit Experiment, unless the problem specifically mentions that the slits have different widths or different intensities, it is always assumed that they are identical coherent sources, meaning $I_1 = I_2 = I_0$.

Q: Why is the maximum intensity $4I_0$ and not just $2I_0$?

This is the magic of constructive interference! Light waves add up their *amplitudes*, not just their intensities. If each slit has amplitude $A$, the max amplitude is $A + A = 2A$. Since intensity is proportional to amplitude squared, the max intensity is proportional to $(2A)^2 = 4A^2$, which is $4I_0$.

Q: I got $3K/4$ as the answer. Why is it wrong?

$3K/4$ is absolutely correct physically! However, the question specifically asks for the value of $n$ in the expression $nK/12$. You just need to find an equivalent fraction by multiplying the top and bottom of $3/4$ by $3$ to get $9/12$. Thus, $n=9$.

Q: What would the intensity be if the path difference was $\lambda/2$?

A path difference of $\lambda/2$ corresponds to a phase difference of $\pi$ (180 degrees). This is the condition for perfect destructive interference. The waves cancel each other out completely, and the intensity would be exactly zero.

Q: Can I use this formula if the light sources are incoherent?

No. For incoherent sources (like two separate light bulbs), there is no stable phase relationship. The interference term $2\sqrt{I_1 I_2}\cos\Delta\phi$ averages out to zero over time. The total intensity would simply be the sum of individual intensities: $I = I_1 + I_2$ everywhere on the screen.

Question

A Young's double slit experiment is performed using monochromatic light of wavelength $\lambda$. The intensity of light at a point on the screen, where the path difference is $\lambda$, is $K$ units. The intensity of light at a point where the path difference is $\frac{\lambda}{6}$ is given by $\frac{nK}{12}$, where $n$ is an integer. The value of $n$ is ......... .

Sigma Percentile · Accepted Answer

### The Symphony of Interference

Imagine standing in a dark room, watching the beautiful, alternating bright and dark bands of a Young's Double Slit Experiment (YDSE) on a screen. This pattern isn't magic; it's pure mathematics and physics dancing together. The intensity of light at any given point on that screen is entirely dictated by the **phase difference** between the two light waves arriving there.

When two coherent waves interfere, their resultant intensity is given by the master equation:

$$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos(\Delta\phi)$$

In a standard YDSE, we assume the slits are identical, meaning they emit light of the same intensity, let's call it $I_0$. So, $I_1 = I_2 = I_0$. Plugging this into our master equation, it beautifully simplifies to:

$$I = 4I_0 \cos^2\left(\frac{\Delta\phi}{2}ight)$$

This simplified equation is our primary tool for this problem.

### Decoding the First Clue

The problem gives us a crucial piece of information: at a point where the path difference $\Delta x$ is $\lambda$, the intensity is $K$. But our intensity formula uses phase difference $\Delta\phi$. How do we bridge this gap? We use the fundamental relationship between phase and path difference:

$$\Delta\phi = \frac{2\pi}{\lambda} \Delta x$$

Substituting $\Delta x = \lambda$ into this relation, we find the phase difference:

$$\Delta\phi = \frac{2\pi}{\lambda} 	imes \lambda = 2\pi$$

Now, let's feed this phase difference back into our simplified intensity formula:

$$I = 4I_0 \cos^2\left(\frac{2\pi}{2}ight) = 4I_0 \cos^2(\pi)$$

Since $\cos(\pi) = -1$, squaring it gives $1$. Therefore, the intensity at this point is $4I_0$. The problem states this intensity is $K$. This gives us our golden key:

$$4I_0 = K \implies I_0 = \frac{K}{4}$$

We have successfully expressed the individual slit intensity $I_0$ in terms of the given constant $K$.

### The Final Act

Now, we are asked to find the intensity at a new point where the path difference is $\frac{\lambda}{6}$. Let's repeat our process. First, find the new phase difference $\Delta\phi'$:

$$\Delta\phi' = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{6} = \frac{\pi}{3}$$

Next, substitute this new phase difference into the intensity formula to find the new intensity $I'$:

$$I' = 4I_0 \cos^2\left(\frac{\pi/3}{2}ight) = 4I_0 \cos^2\left(\frac{\pi}{6}ight)$$

We know that $\cos\left(\frac{\pi}{6}ight)$ (which is $\cos(30^\circ)$) is $\frac{\sqrt{3}}{2}$. Squaring this gives $\frac{3}{4}$.

$$I' = 4I_0 \left(\frac{3}{4}ight) = 3I_0$$

We are almost there! We need the answer in terms of $K$. Let's use our golden key, $I_0 = \frac{K}{4}$:

$$I' = 3\left(\frac{K}{4}ight) = \frac{3K}{4}$$

The problem states this intensity is in the format $\frac{nK}{12}$. To match this format, we simply multiply the numerator and denominator of our result by $3$:

$$I' = \frac{3K 	imes 3}{4 	imes 3} = \frac{9K}{12}$$

By directly comparing $\frac{9K}{12}$ with $\frac{nK}{12}$, it is crystal clear that **$n = 9$**.

Enter Numerical Value:

Visualized Solution

$I = I_{1} + I_{2} + 2 I_{1} I_{2} cos (Δ ϕ)$

$Δ ϕ = \frac{2 π}{λ} Δ x$

$Δ x = λ ⟹ Δ ϕ = 2 π$

$I = K ⟹ 4 I_{0} = K$

$Δ x = \frac{λ}{6} ⟹ Δ ϕ = \frac{π}{3}$

$I^{'} = 4 I_{0} cos^{2} (\frac{π /3}{2})$

$I^{'} = 3 (\frac{K}{4}) = \frac{9 K}{12} ⟹ n = 9$

$What if I_{1} \neq = I_{2} ?$

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

The Symphony of Interference

Decoding the First Clue

The Final Act

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=I1​+I2​+2I1​I2​​cos(Δϕ)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δϕ=λ2π​Δx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δx=λ⟹Δϕ=2π

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=K⟹4I0​=K

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δx=6λ​⟹Δϕ=3π​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I′=4I0​cos2(2π/3​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I′=3(4K​)=129K​⟹n=9

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if I1​=I2​?