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JEE Advanced 2009Paper 2
Exam
JEE Advanced
Year
2009
Paper
Paper 2
Subject
Physics
Chapter
Wave Optics

JEE Advanced 2009 · Paper 2 · Physics · Wave Optics

Column I shows four situations of standard Young's double slit...

Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits S1S_1 and S2S_2. In each of these cases S1P0=S2P0S_1 P_0=S_2 P_0, S1P1−S2P1=λ4S_1 P_1-S_2 P_1=\frac{\lambda}{4} and S1P2−S2P2=λ3S_1 P_2-S_2 P_2=\frac{\lambda}{3}, where λ\lambda is the wavelength of the light used. In the cases B,C\mathrm{B}, \mathrm{C} and D\mathrm{D}, a transparent sheet of refractive index μ\mu and thickness tt is pasted on slit S2S_2. The thickness of the sheets are different in different cases. The phase difference between the light waves reaching a point PP on the screen from the two slits is denoted by δ(P)\delta(P) and the intensity by I(P)I(P). Match each situation given in Column-I with the statement(s) in Column-II valid for that situation. Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the

Options

  1. A(A) q,t, (B) q, (C) t, (D) r,s,t
  2. B(A) q,r,s, (B) r, (C) q,r, (D) q,s,t
  3. C(A) p,s, (B) r, (C) q,t, (D) r,s
  4. D(A) p,s, (B) q, (C) t, (D) r,s,t

Correct answer

D. (A) p,s, (B) q, (C) t, (D) r,s,t

Step-by-step solution

(A) →(p,s)→\rightarrow(\mathrm{p}, \mathrm{s}) \rightarrow Intensity at P0P_0 is maximum. It will continuously decrease from P0P_0 towards P2P_2. (B)→(q)→(B) \rightarrow(q) \rightarrow Path difference due to slap will be compensated by geometrical path difference. Hence, δ(P1)=0\delta\left(P_1\right)=0. (C)→(t)→δ(P0)=λ2,δ(P1)=λ2−λ4=λ4(\mathrm{C}) \rightarrow(\mathrm{t}) \rightarrow \delta\left(P_0\right)=\frac{\lambda}{2}, \delta\left(P_1\right)=\frac{\lambda}{2}-\frac{\lambda}{4}=\frac{\lambda}{4} and δ(P2)=λ2−λ3=λ6\delta\left(P_2\right)=\frac{\lambda}{2}-\frac{\lambda}{3}=\frac{\lambda}{6}. When path difference increases from 0 to λ2\frac{\lambda}{2}, intensity will decrease from maximum to zero. Hence, in this case, I(P2)>I(P1)>I(P0)I\left(P_2\right)>I\left(P_1\right)>I\left(P_0\right) (D)→(r,s,t)(\mathrm{D}) \rightarrow(\mathrm{r}, \mathrm{s}, \mathrm{t}) δ(P0)=3λ4,δ(P1)=3λ4−λ4=λ2\delta\left(P_0\right)=\frac{3 \lambda}{4}, \delta\left(P_1\right)=\frac{3 \lambda}{4}-\frac{\lambda}{4}=\frac{\lambda}{2} and δ(P2)=3λ4−λ3=5λ12\delta\left(P_2\right)=\frac{3 \lambda}{4}-\frac{\lambda}{3}=\frac{5 \lambda}{12} In this case I(P1)=0I\left(P_1\right)=0
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