JEE Challenger
More from Wave Optics
PhyHardAdv 2016 • Paper 2

Intensity Pattern in Youngs Double Slit Experiment Variant

In a modified Young's double slit setup, two small pinholes acting as coherent point sources (S1,S2S_1, S_2) with wavelength 600 nm600\text{ nm} are placed in an opaque sheet lying in the x−yx-y plane. A observing screen is placed parallel to the x−zx-z plane (for z>0z > 0) at a distance D=3 mD = 3\text{ m} from the midpoint of S1S2S_1 S_2, as shown schematically in the figure. The separation distance between the sources is d=0.6003 mmd = 0.6003\text{ mm}. The origin OO lies at the intersection of the screen and the line joining S1S2S_1 S_2. Which of the following assertions is(are) true regarding the intensity pattern formed on the screen?

Question Diagram 1

Options

A

Straight bright and dark fringes parallel to the xx-axis

B

The region very close to the point OO will be dark

Correct
C

Hyperbolic bright and dark fringes with foci symmetrically placed about OO in the xx-direction

D

Semi circular bright and dark fringes centered at point OO

Correct

Topics & Concepts

Wave OpticsInterferenceYoung's Double Slit

Step-by-Step Solution

To determine the nature and characteristics of the intensity pattern formed on the screen, we analyze the setup using a 3D Cartesian coordinate system.

1. Geometry of the Setup and Fringe Shape

Let the midpoint of the line segment S1S2S_1 S_2 be the origin (0,0,0)(0, 0, 0).

  • The line joining the two coherent sources S1S_1 and S2S_2 lies along the yy-axis. Therefore, the coordinates of the sources are: S1=(0,−d2,0)andS2=(0,d2,0)S_1 = \left(0, -\frac{d}{2}, 0\right) \quad \text{and} \quad S_2 = \left(0, \frac{d}{2}, 0\right)
  • The observing screen is parallel to the x−zx-z plane located at y=D=3 my = D = 3\text{ m}, restricted to z>0z > 0.
  • The point OO is the intersection of the screen and the yy-axis, so its coordinates are O=(0,D,0)O = (0, D, 0).

For any point P(x,D,z)P(x, D, z) on the screen (z>0z > 0), the distances S1PS_1 P and S2PS_2 P are: S1P=x2+(D+d2)2+z2=(x2+z2)+(D+d2)2S_1 P = \sqrt{x^2 + \left(D + \frac{d}{2}\right)^2 + z^2} = \sqrt{\left(x^2 + z^2\right) + \left(D + \frac{d}{2}\right)^2} S2P=x2+(D−d2)2+z2=(x2+z2)+(D−d2)2S_2 P = \sqrt{x^2 + \left(D - \frac{d}{2}\right)^2 + z^2} = \sqrt{\left(x^2 + z^2\right) + \left(D - \frac{d}{2}\right)^2}

Let r=x2+z2r = \sqrt{x^2 + z^2} be the radial distance of point PP from the point O(0,D,0)O(0, D, 0) on the screen. The path difference Δx\Delta x between the waves arriving at point PP is: Δx=S1P−S2P=r2+(D+d2)2−r2+(D−d2)2\Delta x = S_1 P - S_2 P = \sqrt{r^2 + \left(D + \frac{d}{2}\right)^2} - \sqrt{r^2 + \left(D - \frac{d}{2}\right)^2}

Notice that Δx\Delta x depends solely on r=x2+z2r = \sqrt{x^2 + z^2}. For constant path difference (and hence uniform intensity), rr must be constant: x2+z2=r2=constantx^2 + z^2 = r^2 = \text{constant}

Since z>0z > 0, this equation represents concentric semicircles centered at point OO. Thus, the intensity pattern consists of semi-circular bright and dark fringes centered at point OO. This confirms Option D.

2. Intensity at the Region Near Point OO

At the point O(0,D,0)O(0, D, 0), we have r=0r = 0. The path difference at OO simplifies to: ΔxO=S1O−S2O=(D+d2)−(D−d2)=d\Delta x_O = S_1 O - S_2 O = \left(D + \frac{d}{2}\right) - \left(D - \frac{d}{2}\right) = d

Given values:

  • d=0.6003 mm=6.003×10−4 md = 0.6003\text{ mm} = 6.003 \times 10^{-4}\text{ m}
  • λ=600 nm=6.00×10−7 m\lambda = 600\text{ nm} = 6.00 \times 10^{-7}\text{ m}

Calculating the ratio of path difference to wavelength: ΔxOλ=6.003×10−4 m6.00×10−7 m=1000.5\frac{\Delta x_O}{\lambda} = \frac{6.003 \times 10^{-4}\text{ m}}{6.00 \times 10^{-7}\text{ m}} = 1000.5

Therefore, the path difference at OO is: ΔxO=(1000+12)λ=(2n+1)λ2(where n=1000)\Delta x_O = \left(1000 + \frac{1}{2}\right) \lambda = (2n + 1)\frac{\lambda}{2} \quad (\text{where } n = 1000)

This corresponds precisely to the condition for destructive interference. Consequently, the region very close to the point OO will be dark. This confirms Option B.

Final Conclusion

  • Option B is CORRECT.
  • Option D is CORRECT.
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