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In Young's double-slit experiment, the separation between the slits is d=0.25 cm and the distance of the screen from the slits is D=100 cm. When the wavelength of light used in the experiment is λ=6000 Å, the intensity at a distance x=4×10-5 m from the central maximum is pqI0, where I0 is the intensity of central maxima and p and q are the smallest integers. What is the value of p+q?

PhysicsWave OpticsJEE Main
Solution:
1342 Upvotes Verified Answer
The correct answer is: 7

Path diff.=xdD

⇒ Path diff.=4×10-5×0.25×10-21

Path diff. = 1 × 1 0 - 7

Phase diff.=path diff.λ×2π

⇒ Phase diff.=1×10-76×10-7×2π

Phase diff.=2π6 ⇒ Phase diff.=π3 ⇒ ϕ=60

IR=I1+I2+2I1I2cos60 ⇒ IR=I1+I2+2I×12

IR=3I ⇒ IR=3I04

p=3, q=4p+q=7

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