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 ce

Question

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

TestHub · Accepted Answer

Due to the induction of the glass plate, the change in path difference is $(μ_{2} - μ_{1}) t$ .

$∵$ 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 $5 λ$ .

$(μ_{2} - μ_{1}) t = 5 λ$

$t = \frac{5 λ}{μ_{2} - μ_{1}} = \frac{5 \times 480 \times 10^{- 9}}{1.7 - 1.4}$

$= 8$ micrometre.

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