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 refractive index 1.7. The point on the screen where the central maximum fell before the glass plates were inserted is now occupied by what had been the fifth bright fringe before. Assume the plates to have the same thickness t and l = 480nm. Then the value of t is

Question

Infinity Learn · Accepted Answer

path difference introduced by  introducing a plate is μ$$-1t$$. here net path difference introduced  is $$1.7-1t$$ $$-1.4-1t$$ $$=5$$λ   as $$5$$ fringes has $$shifted0.3t=5$$ $$x480$$ $$nmt=5$$ x $$1600nm$$ $$=8$$ μm

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