A point source S emitting light of wavelength 600 nm is placed at a very small height h above a flat

Q: Why are the fringes circular and not straight lines like in YDSE?

In YDSE, we have two parallel slit sources, which gives cylindrical wavefronts and straight fringes near the center. Here, we have a point source and its virtual image, which act as two coherent point sources. The symmetry is around the line joining them, leading to circular fringes on a screen perpendicular to this line.

Q: Why is the intensity of the reflected ray only 36%?

Real mirrors are not perfectly reflecting. Some light is absorbed or transmitted. The problem states that only 36% of the incident energy is reflected back.

Q: Why do we use $(n - 1/2)\lambda$ for a maximum instead of $n\lambda$?

When the light reflects off the mirror, it hits a denser medium and undergoes a phase shift of $\pi$ (or $180^\circ$). This phase shift is equivalent to an extra path difference of $\lambda/2$. Therefore, the condition for constructive interference shifts from $n\lambda$ to $(n - 1/2)\lambda$.

Q: Why is the geometric path difference exactly $2h$ at point P?

Point P is directly above the source S. The direct ray travels straight up, covering a distance $D - h$. The reflected ray travels down to the mirror (distance $h$) and then up to the screen (distance $D$), covering a total distance of $D + h$. The difference is $(D + h) - (D - h) = 2h$.

Q: Why does the path difference increase by exactly $\lambda$ for the next maximum?

The condition for consecutive maxima is that their path differences must differ by exactly one full wavelength $\lambda$. When the mirror is shifted, the path difference increases, and to hit the *very next* bright fringe at the same point P, the path difference must have increased by exactly $\lambda$.

Question

A point source $S$ emitting light of wavelength $600 	ext{ nm}$ is placed at a very small height $h$ above a flat reflecting surface $AB$ (see figure). The intensity of the reflected light is $36\%$ of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance $D$ from it.
(a) What is the shape of the interference fringes on the screen?
(b) Calculate the ratio of the minimum to the maximum intensities in the interference fringes formed near the point $P$ (shown in the figure).
(c) If the intensity at point $P$ corresponds to a maximum, calculate the minimum distance through which the reflecting surface $AB$ should be shifted so that the intensity at $P$ again becomes maximum.

Sigma Percentile · Accepted Answer

This problem is a beautiful variation of the classic **Lloyd's Mirror** experiment. It tests your understanding of spatial symmetry, intensity ratios, and the subtle phase shifts that occur when light reflects off a denser medium. Let's break it down step-by-step.

### The Setup: A Mirror and a Source

Imagine a point source $S$ hovering just above a flat mirror $AB$. A screen is placed far away, parallel to the mirror. Light from the source reaches the screen in two ways: directly from the source, and after bouncing off the mirror. These two waves interfere, creating a pattern of bright and dark fringes on the screen.

### Part A: The Geometry of the Fringes

What shape do these fringes take? To answer this, we need to look at the symmetry of the setup. Imagine a vertical axis passing straight through the point source $S$. If you were to rotate the entire setup around this axis, nothing would change. The physical arrangement is perfectly symmetric.

Because of this cylindrical symmetry, the locus of all points on the screen that share the exact same path difference must be a circle centered on that vertical axis. Therefore, the interference fringes observed on the screen will be **circular**.

### Part B: The Battle of Intensities

Next, we need to find the ratio of the minimum to the maximum intensity in the interference pattern. We are given that the reflected light has an intensity that is $36\%$ of the incident light.

Let the intensity of the direct ray be $I_1 = I_0$. 
The intensity of the reflected ray is $I_2 = 0.36 I_0$.

Intensity is proportional to the square of the amplitude ($I \propto a^2$). To find the ratio of the amplitudes, we take the square root of the intensity ratio:

$$\frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{1}{0.36}} = \frac{1}{0.6} = \frac{5}{3}$$

The maximum intensity occurs when the waves interfere constructively ($I_{	ext{max}} \propto (a_1 + a_2)^2$), and the minimum intensity occurs when they interfere destructively ($I_{	ext{min}} \propto (a_1 - a_2)^2$).

Taking the ratio:

$$\frac{I_{	ext{min}}}{I_{	ext{max}}} = \frac{(a_1 - a_2)^2}{(a_1 + a_2)^2} = \frac{(5 - 3)^2}{(5 + 3)^2} = \frac{2^2}{8^2} = \frac{4}{64} = \frac{1}{16}$$

### Part C: The Phase Shift Catch

Now, let's focus on point $P$, which lies on the screen directly above the source $S$. We need to calculate the path difference between the two waves arriving at $P$.

The direct ray travels straight up, covering a distance of $D - h$. 
The reflected ray travels down to the mirror (distance $h$) and then bounces up to the screen (distance $D$), covering a total distance of $D + h$.

The geometric path difference is:

$$\Delta x_{	ext{geo}} = (D + h) - (D - h) = 2h$$

**Here is the crucial catch:** When light reflects off a denser medium (like a mirror), it undergoes an abrupt phase change of $\pi$ radians. This phase shift is mathematically equivalent to an extra path difference of $\lambda/2$.

Because of this phase shift, the condition for constructive interference (a maximum) is no longer $n\lambda$. Instead, it becomes:

$$2h = \left(n - \frac{1}{2}ight)\lambda \quad \dots (i)$$

### Part D: Shifting the Mirror

The problem states that the mirror is shifted by a minimum distance $x$ so that point $P$ becomes a maximum again. If we shift the mirror downwards by $x$, the new height of the source from the mirror becomes $h + x$.

The new geometric path difference at $P$ is now $2(h + x) = 2h + 2x$.

For $P$ to be a maximum *again*, the path difference must have increased to the very next available condition for constructive interference. This means we increment our integer $n$ by 1:

$$2h + 2x = \left(n + 1 - \frac{1}{2}ight)\lambda \quad \dots (ii)$$

### The Final Calculation

We now have a simple system of equations. By subtracting equation (i) from equation (ii), the $2h$ terms cancel out beautifully:

$$2x = \lambda$$

$$x = \frac{\lambda}{2}$$

Given that the wavelength $\lambda$ is $600 	ext{ nm}$, the required shift is:

$$x = \frac{600 	ext{ nm}}{2} = 300 	ext{ nm}$$

And there we have it! A brilliant interplay of geometry, wave optics, and phase shifts.

Visualized Solution

$Experimental Setup$

$Shape of Fringes$

$Intensities of Rays$

$Ratio of Amplitudes$

$Ratio of Minimum to Maximum Intensity$

$Path Difference at P$

$Phase Change on Reflection$

$Initial Condition for Maximum$

$Mirror Shifted by x$

$New Condition for Maximum$

$Calculating the Shift x$

$The Way Forward$

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

The Setup

Part A

Part B

Part C

Part D

The Final Calculation

Similar Questions

Visualized Solution

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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Minimum to Maximum Intensity

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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mirror Shifted by x

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