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In Young's double slit experiment, the slits are 2mm apart are illuminated by photons of two wavelengths I1=12000藲A and I2=10000藲A. at what minimum distance from the common central bright fringe on the screen 2m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

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Solution

Verified by Toppr

Now, from the question we can infer that

n1λ1=n2λ2

so,

n1n2=λ2λ1

or

n1n2=10000A12000A

thus,we have

n1n2=56

5th and 6th fringes will coincide respectively.

the minimum distance is given as

Xmin=n1λ1Dd

here,

n1=5

D = 2m

d = 2mm = 2×10−3m

so,

Xmin=[5×12000×10−10×2]2×10−3

thus, we get

Xmin = 6mm

hence B option is the correct answer

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