A thin flake of transparent material ( n = 1.55) is used to cover one slit of a

Question

A thin flake of transparent material (n = 1.55) is used to cover one slit of a double-slit interference arrangement. The central point on the screen is now occupied by what had been the 4th bright side fringe (m = 4). If ? = 460 nm, what is the thickness of the flake in meters?

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Question 2 A thin flake of transparent material (n -1.55) is used to cover one slit of a double-slit interference arrangement. The central point on the screen is now occupied by what had been the 4th bright side fringe (m = 4). If = 460 nm, what is the thickness of the flake in meters? Number| Units

Explanation / Answer

With no mica fourth bright fringe corresponds to path difference = dsin = 4

With one slit covered, there is an additional path difference due to change of ` = /n in the transparent material

Let unknown thickness of material be L

path difference in material compared to when there was no material is [L/ ` - L/ ] = [nL -L] = [1.55L - L] =0.55L to shift 4th bright fringe to center, this must correspond to 4 = 0.55 L

hence L= 4 /(0.55) = 4(460)x10-9/0.55 = 3.35µm

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