JEE Challenger
More from Wave Optics
PhyHardAdv 2009 • Paper 2

Phase Difference and Intensity in Young Double Slit with Thin Sheets

Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits S1\text{S}_1 and S2\text{S}_2. In each of these cases S1P0=S2P0\text{S}_1\text{P}_0 = \text{S}_2\text{P}_0, S1P1−S2P1=λ/4\text{S}_1\text{P}_1 - \text{S}_2\text{P}_1 = \lambda / 4 and S1P2−S2P2=λ/3\text{S}_1\text{P}_2 - \text{S}_2\text{P}_2 = \lambda / 3, where λ\lambda is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index μ\mu and thickness tt is pasted on slit S2\text{S}_2. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P\text{P} on the screen from the two slits is denoted by δ(P)\delta(\text{P}) and the intensity by I(P)I(\text{P}). Match each situation given in Column I with the statement(s) in Column II valid for that situation.

Column I

(A)

Diagram 0

(B) (μ−1)t=λ/4(\mu - 1)t = \lambda / 4

(C) (μ−1)t=λ/2(\mu - 1)t = \lambda / 2

(D) (μ−1)t=3λ/4(\mu - 1)t = 3\lambda / 4

Column II

(p) δ(P0)=0\delta(\text{P}_0) = 0

(q) δ(P1)=0\delta(\text{P}_1) = 0

(r) I(P1)=0I(\text{P}_1) = 0

(s) I(P0)>I(P1)I(\text{P}_0) > I(\text{P}_1)

(t) I(P2)>I(P1)I(\text{P}_2) > I(\text{P}_1)

Options

A

(A)

B

(B) (μ−1)t=λ/4(\mu - 1)t = \lambda / 4

C

(C) (μ−1)t=λ/2(\mu - 1)t = \lambda / 2

D

(D) (μ−1)t=3λ/4(\mu - 1)t = 3\lambda / 4

Topics & Concepts

Wave OpticsYoung's Double SlitInterference

Step-by-Step Solution

To find the correct matching between Column I and Column II, we determine the net optical path difference Δx\Delta x and phase difference δ\delta at points P0\text{P}_0, P1\text{P}_1, and P2\text{P}_2 for each case.

The net optical path difference between the waves arriving at any point P\text{P} on the screen from the two slits is given by:
Δx(P)=(S1P−S2P)−(μ−1)t\Delta x(\text{P}) = (\text{S}_1\text{P} - \text{S}_2\text{P}) - (\mu - 1)t
where the sheet of thickness tt and refractive index μ\mu is introduced in front of slit S2\text{S}_2.

The corresponding phase difference is:
δ(P)=2πλ∣Δx(P)∣\delta(\text{P}) = \frac{2\pi}{\lambda} |\Delta x(\text{P})|
and the resulting intensity is:
I(P)=Imax⁡cos⁡2(δ(P)2)I(\text{P}) = I_{\max} \cos^2\left(\frac{\delta(\text{P})}{2}\right)

Given geometrical path differences:

  • At P0\text{P}_0: S1P0−S2P0=0\text{S}_1\text{P}_0 - \text{S}_2\text{P}_0 = 0
  • At P1\text{P}_1: S1P1−S2P1=λ4\text{S}_1\text{P}_1 - \text{S}_2\text{P}_1 = \frac{\lambda}{4}
  • At P2\text{P}_2: S1P2−S2P2=λ3\text{S}_1\text{P}_2 - \text{S}_2\text{P}_2 = \frac{\lambda}{3}

Case (A): No sheet is placed, (μ−1)t=0(\mu - 1)t = 0

  • At P0\text{P}_0:
    Δx(P0)=0  ⟹  δ(P0)=0\Delta x(\text{P}_0) = 0 \implies \delta(\text{P}_0) = 0
    I(P0)=Imax⁡cos⁡2(0)=Imax⁡I(\text{P}_0) = I_{\max}\cos^2(0) = I_{\max}
  • At P1\text{P}_1:
    Δx(P1)=λ4  ⟹  δ(P1)=2πλ(λ4)=π2\Delta x(\text{P}_1) = \frac{\lambda}{4} \implies \delta(\text{P}_1) = \frac{2\pi}{\lambda}\left(\frac{\lambda}{4}\right) = \frac{\pi}{2}
    I(P1)=Imax⁡cos⁡2(π4)=Imax⁡2I(\text{P}_1) = I_{\max}\cos^2\left(\frac{\pi}{4}\right) = \frac{I_{\max}}{2}
  • At P2\text{P}_2:
    Δx(P2)=λ3  ⟹  δ(P2)=2πλ(λ3)=2π3\Delta x(\text{P}_2) = \frac{\lambda}{3} \implies \delta(\text{P}_2) = \frac{2\pi}{\lambda}\left(\frac{\lambda}{3}\right) = \frac{2\pi}{3}
    I(P2)=Imax⁡cos⁡2(π3)=Imax⁡4I(\text{P}_2) = I_{\max}\cos^2\left(\frac{\pi}{3}\right) = \frac{I_{\max}}{4}

Checking the statements:

  • δ(P0)=0\delta(\text{P}_0) = 0 is true   ⟹  (p)\implies \mathbf{(p)}
  • I(P0)>I(P1)I(\text{P}_0) > I(\text{P}_1) since Imax⁡>Imax⁡2I_{\max} > \frac{I_{\max}}{2} is true   ⟹  (s)\implies \mathbf{(s)}

Hence, (A) →\to (p), (s).

Case (B): (μ−1)t=λ4(\mu - 1)t = \frac{\lambda}{4}

  • At P0\text{P}_0:
    Δx(P0)=0−λ4=−λ4  ⟹  δ(P0)=π2\Delta x(\text{P}_0) = 0 - \frac{\lambda}{4} = -\frac{\lambda}{4} \implies \delta(\text{P}_0) = \frac{\pi}{2}
  • At P1\text{P}_1:
    Δx(P1)=λ4−λ4=0  ⟹  δ(P1)=0\Delta x(\text{P}_1) = \frac{\lambda}{4} - \frac{\lambda}{4} = 0 \implies \delta(\text{P}_1) = 0
    I(P1)=Imax⁡cos⁡2(0)=Imax⁡I(\text{P}_1) = I_{\max}\cos^2(0) = I_{\max}
  • At P2\text{P}_2:
    Δx(P2)=λ3−λ4=λ12  ⟹  δ(P2)=π6\Delta x(\text{P}_2) = \frac{\lambda}{3} - \frac{\lambda}{4} = \frac{\lambda}{12} \implies \delta(\text{P}_2) = \frac{\pi}{6}
    I(P2)=Imax⁡cos⁡2(π12)<Imax⁡=I(P1)I(\text{P}_2) = I_{\max}\cos^2\left(\frac{\pi}{12}\right) < I_{\max} = I(\text{P}_1)

Checking the statements:

  • δ(P1)=0\delta(\text{P}_1) = 0 is true   ⟹  (q)\implies \mathbf{(q)}

Hence, (B) →\to (q).

Case (C): (μ−1)t=λ2(\mu - 1)t = \frac{\lambda}{2}

  • At P0\text{P}_0:
    Δx(P0)=0−λ2=−λ2  ⟹  δ(P0)=π\Delta x(\text{P}_0) = 0 - \frac{\lambda}{2} = -\frac{\lambda}{2} \implies \delta(\text{P}_0) = \pi
    I(P0)=Imax⁡cos⁡2(π2)=0I(\text{P}_0) = I_{\max}\cos^2\left(\frac{\pi}{2}\right) = 0
  • At P1\text{P}_1:
    Δx(P1)=λ4−λ2=−λ4  ⟹  δ(P1)=π2\Delta x(\text{P}_1) = \frac{\lambda}{4} - \frac{\lambda}{2} = -\frac{\lambda}{4} \implies \delta(\text{P}_1) = \frac{\pi}{2}
    I(P1)=Imax⁡cos⁡2(π4)=Imax⁡2I(\text{P}_1) = I_{\max}\cos^2\left(\frac{\pi}{4}\right) = \frac{I_{\max}}{2}
  • At P2\text{P}_2:
    Δx(P2)=λ3−λ2=−λ6  ⟹  δ(P2)=π3\Delta x(\text{P}_2) = \frac{\lambda}{3} - \frac{\lambda}{2} = -\frac{\lambda}{6} \implies \delta(\text{P}_2) = \frac{\pi}{3}
    I(P2)=Imax⁡cos⁡2(π6)=34Imax⁡I(\text{P}_2) = I_{\max}\cos^2\left(\frac{\pi}{6}\right) = \frac{3}{4}I_{\max}

Checking the statements:

  • I(P2)>I(P1)I(\text{P}_2) > I(\text{P}_1) since 34Imax⁡>12Imax⁡\frac{3}{4}I_{\max} > \frac{1}{2}I_{\max} is true   ⟹  (t)\implies \mathbf{(t)}

Hence, (C) →\to (t).

Case (D): (μ−1)t=3λ4(\mu - 1)t = \frac{3\lambda}{4}

  • At P0\text{P}_0:
    Δx(P0)=0−3λ4=−3λ4  ⟹  δ(P0)=3π2\Delta x(\text{P}_0) = 0 - \frac{3\lambda}{4} = -\frac{3\lambda}{4} \implies \delta(\text{P}_0) = \frac{3\pi}{2}
    I(P0)=Imax⁡cos⁡2(3π4)=Imax⁡2I(\text{P}_0) = I_{\max}\cos^2\left(\frac{3\pi}{4}\right) = \frac{I_{\max}}{2}
  • At P1\text{P}_1:
    Δx(P1)=λ4−3λ4=−λ2  ⟹  δ(P1)=π\Delta x(\text{P}_1) = \frac{\lambda}{4} - \frac{3\lambda}{4} = -\frac{\lambda}{2} \implies \delta(\text{P}_1) = \pi
    I(P1)=Imax⁡cos⁡2(π2)=0I(\text{P}_1) = I_{\max}\cos^2\left(\frac{\pi}{2}\right) = 0
  • At P2\text{P}_2:
    Δx(P2)=λ3−3λ4=−5λ12  ⟹  δ(P2)=5π6\Delta x(\text{P}_2) = \frac{\lambda}{3} - \frac{3\lambda}{4} = -\frac{5\lambda}{12} \implies \delta(\text{P}_2) = \frac{5\pi}{6}
    I(P2)=Imax⁡cos⁡2(5π12)>0I(\text{P}_2) = I_{\max}\cos^2\left(\frac{5\pi}{12}\right) > 0

Checking the statements:

  • I(P1)=0I(\text{P}_1) = 0 is true   ⟹  (r)\implies \mathbf{(r)}
  • I(P0)>I(P1)I(\text{P}_0) > I(\text{P}_1) since Imax⁡2>0\frac{I_{\max}}{2} > 0 is true   ⟹  (s)\implies \mathbf{(s)}
  • I(P2)>I(P1)I(\text{P}_2) > I(\text{P}_1) since I(P2)>0I(\text{P}_2) > 0 is true   ⟹  (t)\implies \mathbf{(t)}

Hence, (D) →\to (r), (s), (t).

Final Matching:

  • (A) →\to (p), (s)
  • (B) →\to (q)
  • (C) →\to (t)
  • (D) →\to (r), (s), (t)
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