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CBSE Class 12 Physics 2020 Outside Delhi Set 1 Paper

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Question : 33 of 37
Marks: +1, -0
Two coherent light waves of intensity 5×5 \times 10−2 Wm−210^{-2} \, \text{Wm}^{-2} each superimpose and produce the interference pattern on a screen. At a point where the path difference between the waves is λ6,λ\frac{\lambda}{6}, \lambda being wavelength of the wave, find the
(a) phase difference between the waves
(b) resultant intensity at the point
(c) resultant intensity in terms of the intensity at the maximum
Solution:  
(a) Δx=λΔϕ2π\Delta x = \frac{\lambda \Delta \phi}{2 \pi}, where Δx=\Delta x = Path difference and
Δϕ= Phase difference \Delta \phi = \text{ Phase difference }
Δx=λ6\Delta x = \frac{\lambda}{6}
∴Δϕ= phase difference \therefore \Delta \phi = \text{ phase difference } =λ6×2πλ=2π6= \frac{\lambda}{6} \times \frac{2 \pi}{\lambda} = \frac{2 \pi}{6}
(b) Resultant intensity =I=I0cos⁡2Δϕ2= I = I_0 \cos^2 \frac{\Delta \phi}{2}
 Or, I=I0cos⁡2π6\text{ Or, } I = I_0 \cos^2 \frac{\pi}{6}
 Or, I=5×10−2cos⁡2π6\text{ Or, } I = 5 \times 10^{-2} \cos^2 \frac{\pi}{6}
 Or, I=5×10−2(32)2\text{ Or, } I = 5 \times 10^{-2} \left( \frac{\sqrt{3}}{2} \right)^2
∴I=3.75×10−2 Wm−2\therefore I = 3.75 \times 10^{-2} \, \text{Wm}^{-2}
(c) Resultant intensity at maxima =Imax⁡=4I0= I_{\max} = 4 I_0
Resultant intensity =IRESULTANT=3l04= I_{\text{RESULTANT}} = \frac{3 l_0}{4}
(c) Resultant intensity at maxima =Imax⁡=4I0= I_{\max} = 4 I_0
 Resultant intensity =IRESULTANT=3I04\text{ Resultant intensity } = I_{\text{RESULTANT}} = \frac{3 I_0}{4}
IRESULTANTIMAX=3I044I0=316\frac{I_{\text{RESULTANT}}}{I_{\text{MAX}}} = \frac{ \frac{3 I_0}{4} }{4 I_0} = \frac{3}{16}
∴IRESULTANT=316IMAX\therefore I_{\text{RESULTANT}} = \frac{3}{16} I_{\text{MAX}}
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