JEE Advanced 2016

LEVELJEE Advanced

Animated Solution for Physics - Optics: While conducting the Young's double slit experiment, a student replaced the two slits with a large opaque plate in the $x$ - $y$ plane containing two small holes that act as two coherent point sources ( $S_{1}, S_{2}$ ) emitting light of wavelength $600 nm$ . The student mistakenly placed the screen parallel to the $x$ - $z$ plane (for $z > 0$ ) at a distance $D = 3 m$ from the mid-point of $S_{1} S_{2}$ , as shown schematically in the figure. The distance between the sources $d = 0.6003 mm$ . The origin $O$ is at the intersection of the screen and the line joining $S_{1} S_{2}$ .

Q: Why is the path difference at the origin exactly $d$?

Because the origin $O$ lies on the same straight line (the y-axis) that passes through both sources $S_1$ and $S_2$. The wave from the further source must travel the distance between the sources ($d$) plus the distance from the nearer source to $O$. Thus, the difference in their path lengths is exactly $d$.

Q: Why does the problem specify $z > 0$?

The condition $z > 0$ restricts the screen to only the upper half of the $x$-$z$ plane. Because we only have half of the plane, the circular fringes are cut in half, resulting in semi-circles.

Select Answer:

* Multiple Correct

Visualized Solution

$Analyzing the Setup$

$The sources S_{1} and S_{2} lie on the y -axis.$
$The screen is the x - z plane at y = D .$

$Path Difference at Origin O$

$Point O lies on the y -axis.$
$Δ x_{O} = S_{1} O - S_{2} O$

$Calculating Path Difference$

$Δ x_{O} = d = 0.6003 mm$
$Δ x_{O} = 600300 nm$

$Relating to Wavelength$

$λ = 600 nm$
$\frac{Δ x _{O}}{λ} = \frac{600300}{600} = 1000.5$

$Condition for Minima$

$Δ x_{O} = 1000.5 λ = (1000 + \frac{1}{2}) λ$
$Δ x_{O} = (n + \frac{1}{2}) λ ⟹ Dark Fringe$

$Shape of Fringes$

$Locus of constant path difference: Hyperboloid$
$Intersection with screen (⊥ to axis): Circles$
$Since z > 0, fringes are semi-circles.$

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

Solution Diagram

Analyzing the Setup

Imagine you are standing in a lab, setting up a Young's Double Slit Experiment. But instead of the usual setup, you place the screen perpendicular to the line joining the two sources! Let's break down what happens in this unique geometry.

The two coherent sources,

S_{1}

and

S_{2}

, are located on the

y

-axis. The screen is placed parallel to the

x

z

plane at a distance

D = 3 m

. The origin

O

is exactly where the

y

-axis pierces the screen.

The Master Equation

Path Difference at the Origin

To understand the interference pattern, we must first look at the center of our screen, point

O

. What is the path difference between the light waves reaching this point?

Since

O

lies directly on the line passing through

S_{1}

and

S_{2}

, the wave from

S_{1}

has to travel an extra distance exactly equal to the separation between the sources,

d

, compared to the wave from

S_{2}

Therefore, the path difference at the origin is simply

Δ x_{O} = d

Final Calculation

Bright or Dark?

Now, let's plug in the numbers. We are given

d = 0.6003 mm

, which is

600300 nm

. The wavelength of the light is

λ = 600 nm

Let's find how many wavelengths fit into this path difference:

\frac{Δ x _{O}}{λ} = \frac{600300}{600} = 1000.5

This means the path difference is exactly

1000.5 λ

, or

(1000 + \frac{1}{2}) λ

. Because it is an odd multiple of half the wavelength, the waves arrive perfectly out of phase. This results in destructive interference, making the region very close to point

O

completely dark!

The Geometry of the Fringes

Finally, what is the shape of the fringes on the screen? In 3D space, the locus of points that have a constant path difference from two fixed points is a hyperboloid.

When we intersect this hyperboloid with a flat screen that is perpendicular to the axis of the sources, the cross-sections are perfect circles. However, our screen only exists for

z > 0

. Therefore, we will observe semi-circular bright and dark bands centered at

O

Common Traps & FAQs

Why is the path difference at the origin exactly

d

Because the origin

O

lies on the same straight line (the y-axis) that passes through both sources

S_{1}

and

S_{2}

. The wave from the further source must travel the distance between the sources (

d

) plus the distance from the nearer source to

O

. Thus, the difference in their path lengths is exactly

d

Why does a path difference of

1000.5 λ

mean a dark fringe?

A path difference of a half-integer multiple of the wavelength (like

0.5 λ, 1.5 λ, 1000.5 λ

) means the peak of one wave aligns perfectly with the trough of the other. They cancel each other out, causing destructive interference and a dark fringe.

Why are the fringes circular instead of the usual straight lines?

In a standard YDSE, the screen is parallel to the line joining the slits, which approximates the hyperboloid intersections as straight lines. Here, the screen is perpendicular to the line joining the slits. Slicing a hyperboloid perpendicular to its axis always yields concentric circles.

Why does the problem specify