In Young's double slit experiment, the intensity at a point is (1/4) of the maximum intensity. Angul

Q: Why do we use $I = I_{\max} \cos^2(\phi/2)$ instead of just adding intensities?

Because light is a wave! When two coherent light waves meet, their electric fields add up (superposition), not their intensities. The intensity is proportional to the square of the resultant electric field, which leads to the $\cos^2(\phi/2)$ interference term.

Q: Why did we take $\cos(\phi/2) = 1/2$ and not $-1/2$?

Taking $-1/2$ would give $\phi/2 = 2\pi/3$, so $\phi = 4\pi/3$. This corresponds to the next point on the screen where the intensity is $I_{\max}/4$. The question implicitly asks for the *first* such point (closest to the central maximum), which corresponds to the smallest positive phase difference.

Q: What is the difference between phase difference ($\phi$) and path difference ($\Delta x$)?

Path difference is the actual physical extra distance (in meters) one wave travels compared to the other. Phase difference is how much one wave is shifted relative to the other in terms of its wave cycle (in radians). They are linked by the relation $\phi = \frac{2\pi}{\lambda} \Delta x$.

Q: Why is the path difference $\Delta x = d \sin \theta$?

If you draw a perpendicular from the top slit to the ray from the bottom slit, the extra distance the bottom ray travels is the opposite side of a right triangle with hypotenuse $d$ and angle $\theta$. Hence, $\Delta x = d \sin \theta$. This assumes the screen is very far away ($D \gg d$).

Q: Can the angular position $\theta$ be greater than $90^\circ$?

No, in the standard YDSE setup, the screen is a flat plane in front of the slits. The maximum possible angle a ray can make to hit the screen is approaching $90^\circ$ (parallel to the plane of the slits).

Question

In Young's double slit experiment, the intensity at a point is (1/4) of the maximum intensity. Angular position of this point is

Options:

(A) $\sin^{-1}\left(\frac{\lambda}{d}\right)$

(B) $\sin^{-1}\left(\frac{\lambda}{2d}\right)$

(C) $\sin^{-1}\left(\frac{\lambda}{3d}\right)$

(D) $\sin^{-1}\left(\frac{\lambda}{4d}\right)$

Sigma Percentile · Accepted Answer

### Unraveling the Angular Position in Young's Double Slit Experiment

Imagine you are looking at the beautiful, alternating bright and dark fringes on a screen in a Young's Double Slit Experiment. The central maximum is the brightest point, but as you move away, the intensity gradually drops. In this problem, we are hunting for a very specific spot: the point where the intensity has dropped to exactly one-fourth of its maximum value. Our mission is to find the angular position, $	heta$, of this exact location.

### The Master Equation of Interference

To connect the physical position on the screen to the brightness of the light, we need the fundamental equation for interference intensity. When two coherent light waves meet, their resultant intensity $I$ is given by:

$$I = I_{\max} \cos^2\left(\frac{\phi}{2}ight)$$

Here, $I_{\max}$ is the maximum possible intensity (found at the central bright fringe), and $\phi$ is the phase difference between the two light waves arriving at that specific point.

### Finding the Phase Difference

The problem hands us a crucial piece of information: at our mystery point, the intensity $I$ is $\frac{I_{\max}}{4}$. Let's plug this straight into our master equation:

$$\frac{I_{\max}}{4} = I_{\max} \cos^2\left(\frac{\phi}{2}ight)$$

Notice how $I_{\max}$ appears on both sides? We can elegantly cancel it out, leaving us with a pure trigonometric equation:

$$\frac{1}{4} = \cos^2\left(\frac{\phi}{2}ight)$$

Taking the square root of both sides gives us:

$$\cos\left(\frac{\phi}{2}ight) = \frac{1}{2}$$

Now, we ask ourselves: for what angle is the cosine equal to $\frac{1}{2}$? From our standard trigonometric values, we know this happens at $60^\circ$, or $\frac{\pi}{3}$ radians. Therefore:

$$\frac{\phi}{2} = \frac{\pi}{3}$$

Multiplying by 2, we find the phase difference:

$$\phi = \frac{2\pi}{3}$$

### From Phase to Path Difference

We have the phase difference, but phase is an abstract angular measure. We need to translate this into a physical distance. The bridge between phase difference $\phi$ and path difference $\Delta x$ is one of the most important relations in wave optics:

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

Let's substitute our calculated phase difference into this bridge equation:

$$\frac{2\pi}{3} = \frac{2\pi}{\lambda} \Delta x$$

The $2\pi$ terms cancel out perfectly, revealing the physical path difference:

$$\Delta x = \frac{\lambda}{3}$$

This tells us that the light wave from the lower slit travels exactly one-third of a wavelength further than the light wave from the upper slit to reach our point.

### The Geometry of YDSE

Finally, we need to connect this path difference to the angular position $	heta$. If we look at the geometry of the double-slit setup, assuming the screen is far away ($D \gg d$), the path difference is simply the component of the slit separation $d$ along the direction of the rays. This is given by the classic geometric relation:

$$\Delta x = d \sin 	heta$$

Now, we have two expressions for the path difference. Let's equate them:

$$d \sin 	heta = \frac{\lambda}{3}$$

Isolating $\sin 	heta$, we get:

$$\sin 	heta = \frac{\lambda}{3d}$$

To find the angle itself, we take the inverse sine:

$$	heta = \sin^{-1}\left(\frac{\lambda}{3d}ight)$$

And there we have it! By seamlessly connecting intensity to phase, phase to path difference, and path difference to geometry, we've pinpointed the exact angular position.

Animated Solution for Physics - Optics: In Young's double slit experiment, the intensity at a point is (1/4) of the maximum intensity. Angular position of this point is

Select Answer:

Visualized Solution

Visualizing the Setup

The Intensity Formula

Substituting the Given Intensity

Solving for the Phase Angle

Calculating Phase Difference $ϕ$

Relating Phase to Path Difference

Calculating Path Difference $Δ x$

Geometric Path Difference

Final Angular Position

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

Unraveling the Angular Position in Young's Double Slit Experiment

The Master Equation of Interference

Finding the Phase Difference

From Phase to Path Difference

The Geometry of YDSE

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