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Asked by eknathmote50 | 24 May, 2019, 11:48: AM

Expert Answer

Because source S is not at the centre line therefor the centrak fringe does not formed at centre of the screen i.e at o

lets first calculate the shift due to non symmetrical position of S

for centre bright fringe

increment x equals 0

Let us assume that central fringe forms at point R which is below point O

i.e

S S subscript 1 plus S subscript 1 R minus open parentheses S S subscript 2 plus S subscript 2 R close parentheses equals 0 S S subscript 1 minus S S subscript 2 plus S subscript 1 R minus S subscript 2 R equals 0 d minus square root of 2 d plus d S i n theta equals 0 open square brackets S S subscript 2 equals square root of d squared plus d squared end root equals d square root of 2 close square brackets d minus square root of 2 d plus d y over D equals 0 y equals open parentheses square root of 2 minus 1 close parentheses D y subscript 1 equals open parentheses square root of 2 minus 1 close parentheses D

since Y is coming positive therefore our assumtion is correct that central fringe forms at Point R which is below O

Now to bring the central fringe at point O sheet must be introduced in front of

S subscript 1

such that the fringe pattern will shift above R

shift occurs because of mica sheet of refractive index 1.5 is

y subscript 2 equals D over d open parentheses mu minus 1 close parentheses t y subscript 2 equals D over d open parentheses 1.5 minus 1 close parentheses t

This shift must be equal to that of Y1

on equating Y1 and Y2

D over d open parentheses 1.50 minus 1 close parentheses t equals open parentheses square root of 2 minus 1 close parentheses D t equals 2 open parentheses square root of 2 minus 1 close parentheses d

Answered by Utkarsh Lokhande | 24 May, 2019, 03:18: PM

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