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CBSE Class 12 Physics 2022 Term 2 Outside Delhi Set 1 Solved Paper

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Question : 10 of 12
Marks: +1, -0
In a Young's double slit experiment using light of wavelength 600 nm600 \text{ nm}, the slit separation is 0.8 mm0.8 \text{ mm} and the screen is kept 1.6 m1.6 \text{ m} from the plane of the slits.
Calculate
(i) the fringe width
(ii) the distance of (a) third minimum and (b) fifth maximum, from the central maximum.3
Solution:  
(i) Fringe width =β=λDd= \beta = \frac{\lambda D}{d}
∴β=600×10−9×1.60.8×10−3=12×10−4 m\therefore \beta = \frac{600 \times 10^{-9} \times 1.6}{0.8 \times 10^{-3}} = 12 \times 10^{-4} \text{ m}
(ii) (a) Distance of 3rd minimum from central maximum =52β=30×10−4 m= \frac{5}{2} \beta = 30 \times 10^{-4} \text{ m}
(b) Distance of 5th maximum from central fringe =5β=60×10−4 m= 5\beta = 60 \times 10^{-4} \text{ m}
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