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Question: Answered & Verified by Expert
One slit of a double-slit experiment is covered by a thin glass plate of refractive index 1.4 , and the other by a thin glass plate of the refractive index 1.7 . The point on the screen where the central maximum fall before the glass plate was inserted, is now occupied by the fifth bright fringe. Assume the plate have the same thickness t and wavelength of light 480 nm , then the value of t is
PhysicsWave OpticsJEE Main
Options:
  • A 2.4μm
  • B 4.8μm
  • C 8μm
  • D 16μm
Solution:
2939 Upvotes Verified Answer
The correct answer is: 8μm
Due to the induction of the glass plate, the change in path difference is μ2-μ1t .

Before inserting the glass plate, the path difference for central maxima is zero.

After introducing glass plate, the change in path difference is equal to 5λ .

μ2-μ1t=5λ

t=5λμ2-μ1=5×480×10-91.7-1.4

=8 micrometre.

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