In a modified Young's double slit experiment, a monochromatic uniform and parallel beam of light of

Q: Why do we use power instead of intensity in the interference formula?

In standard YDSE problems, the slits are identical, so we often just add intensities. However, here the apertures have different areas, meaning the total energy (power) passing through them is different. Since power is proportional to amplitude squared (just like intensity), the interference formula $P = P_1 + P_2 + 2\sqrt{P_1 P_2} \cos \phi$ is perfectly valid and necessary here.

Q: Why does the lens not introduce any extra path difference?

A perfectly symmetric, ideal converging lens focuses parallel rays to its focal point in such a way that all optical paths from a plane perpendicular to the incident rays to the focal point are exactly equal. The lens geometry perfectly compensates for the different geometric distances.

Q: What is the significance of the 10% factor?

When light passes through a finite aperture, it diffracts (spreads out). The 10% factor accounts for the fact that only a fraction of the total energy passing through the aperture remains in the strictly forward direction to be focused exactly at the central focal spot $F$.

Q: Why is the path difference formula $(\mu - 1)t$?

Light travels slower in a medium of refractive index $\mu$. The time it takes to travel a distance $t$ in the medium is the same time it would take to travel a distance $\mu t$ in a vacuum. Therefore, the 'optical path' is $\mu t$. Compared to a ray traveling the same geometric distance $t$ in air (optical path $t$), the extra optical path is $\mu t - t = (\mu - 1)t$.

Q: Can we add the powers directly if the sources were incoherent?

Yes! If the sources were incoherent, the phase difference $\phi$ would fluctuate rapidly randomly. The average value of $\cos \phi$ over time would be zero. The interference term $2\sqrt{P_1 P_2} \cos \phi$ would vanish, and the resultant power would simply be the sum of the individual powers: $P = P_1 + P_2$.

Question

In a modified Young's double slit experiment, a monochromatic uniform and parallel beam of light of wavelength $6000 	ext{ \AA}$ and intensity $(10/\pi) 	ext{ Wm}^{-2}$ is incident normally on two apertures $A$ and $B$ of radii $0.001 	ext{ m}$ and $0.002 	ext{ m}$ respectively. A perfectly transparent film of thickness $2000 	ext{ \AA}$ and refractive index $1.5$ for the wavelength of $6000 	ext{ \AA}$ is placed in front of aperture $A$ (see figure). Calculate the power (in W) received at the focal spot $F$ of the lens. The lens is symmetrically placed with respect to the apertures. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focal spot.

Sigma Percentile · Accepted Answer

This problem is a beautiful amalgamation of several core concepts in wave optics: intensity, power, optical path length, and the interference of waves with unequal amplitudes. Let's break down the journey step-by-step.

### Analyzing the Setup

We are given a modified Young's double-slit setup. Instead of narrow slits, we have two circular apertures, $A$ and $B$, with different radii. A uniform parallel beam of light is incident on them. The first crucial realization is that the light passing through these apertures will have different total powers because their areas are different.

Intensity $I$ is defined as power per unit area. Therefore, the power $P$ received by an aperture of radius $r$ is:

$P = I 	imes \pi r^2$

Let's calculate the power received by each aperture. For aperture $A$ ($r_A = 0.001 	ext{ m}$):

$P_A = \left(\frac{10}{\pi}ight) 	imes \pi (0.001)^2 = 10^{-5} 	ext{ W}$

For aperture $B$ ($r_B = 0.002 	ext{ m}$):

$P_B = \left(\frac{10}{\pi}ight) 	imes \pi (0.002)^2 = 4 	imes 10^{-5} 	ext{ W}$

### The Effective Power at the Focal Spot

The problem states that only $10\%$ of the power received by each aperture goes in the original direction and is brought to the focal spot $F$ by the lens. This accounts for diffraction and scattering effects. So, the actual powers of the two interfering beams at $F$ are:

$P_1 = 10\% 	ext{ of } P_A = 10^{-6} 	ext{ W}$

$P_2 = 10\% 	ext{ of } P_B = 4 	imes 10^{-6} 	ext{ W}$

### The Optical Path Difference

Next, we must account for the transparent film placed in front of aperture $A$. When light travels through a medium of refractive index $\mu$ and thickness $t$, it covers an optical path of $\mu t$. The equivalent path in a vacuum (or air) would just be $t$. Therefore, the extra path length introduced by the film is:

$\Delta x = (\mu - 1)t$

Substituting the given values ($\mu = 1.5$, $t = 2000 	ext{ \AA}$):

$\Delta x = (1.5 - 1) 	imes 2000 	ext{ \AA} = 1000 	ext{ \AA}$

### The Phase Difference

To use the interference formula, we need the phase difference $\phi$ between the two waves arriving at $F$. The phase difference is directly proportional to the path difference:

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

Plugging in our values ($\lambda = 6000 	ext{ \AA}$):

$\phi = \frac{2\pi}{6000} 	imes 1000 = \frac{\pi}{3} 	ext{ rad}$

### Final Calculation: The Resultant Power

Finally, the two beams interfere at the focal spot $F$. Since the beams have different powers (and thus different amplitudes), we use the general interference formula for resultant power:

$P = P_1 + P_2 + 2\sqrt{P_1 P_2} \cos \phi$

Let's substitute everything we've found:

$P = 10^{-6} + 4 	imes 10^{-6} + 2\sqrt{(10^{-6})(4 	imes 10^{-6})} \cos \left(\frac{\pi}{3}ight)$

Simplifying the square root and the cosine term ($\cos(\pi/3) = 1/2$):

$P = 5 	imes 10^{-6} + 2(2 	imes 10^{-6}) 	imes \frac{1}{2}$

$P = 5 	imes 10^{-6} + 2 	imes 10^{-6} = 7 	imes 10^{-6} 	ext{ W}$

And there we have it! The resultant power received at the focal spot is **$7 	imes 10^{-6} 	ext{ W}$**.

Visualized Solution

$Visualizing the Setup$

$Power Received by Apertures$

$Substituting Values for P_{A} and P_{B}$

$Calculating P_{A} and P_{B}$

$Power Reaching Focal Spot F$

$Calculating P_{1} and P_{2}$

$Path Difference Due to Film$

$Substituting Values for Δ x$

$Calculating Δ x$

$Phase Difference$

$Substituting Values for ϕ$

$Calculating ϕ$

$Resultant Power Formula$

$Substituting Values for Resultant Power$

$Final Calculation$

$The Way Forward$

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

Analyzing the Setup

The Effective Power at the Focal Spot

The Optical Path Difference

The Phase Difference

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; } Power Received by Apertures

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for PA​ and PB​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating PA​ and PB​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Power Reaching Focal Spot F

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating P1​ and P2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Path Difference Due to Film

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for Δx

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for ϕ

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resultant Power Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for Resultant Power

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

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