The arrangement for a mirror experiment is shown in figure. S is a point source of frequency `6xx10^(14)Hz`. D and C represent the two ends of a mirror placed horizontally and LOM represents the screen.

Find the fringe width of the fringe pattern ?

Options

A. 0.05 cm
B. 0.25 cm
C. 0.01 cm
D. 0.1 cm

Answer

a

Explanation

Fringes will be observed in the region between `P_(1)` and `P_(2)` becasue the reflected rays lie only in this region
From similar triangles BDS' and `S P_(2)A`,
`(AP_(2))/(BS') = (AS')/(BD)`
`:. AP_(2) = ((AS')(BS'))/(BD)`
`= ((190 + 5 + 5)(0.1))/(5) = 4` cm
Similarly, in triangle BCS' and `S' P _(1) A`,
`(AP_(1))/(BS') = (AS')/(BD)`
`:. ((AS")(BS"))/(BC) = ((190 + 5 +5)(0.1))/(10) = 2` cm
`:. P_(1) P_(2) = AP_(2) - AP_(1) = 2` cm
Wavelength of the light,
`lambda = (c)/(f) = (3 xx 10^(8))/(6 xx 10^(14)) = 5 xx 10^(-7)` m
Fringe width, `beta = (lambda D)/(d)`
Here,` D= S' A = 190 + 5 +5 = 200 cm = 2.0 m, d = SS' = 2 mm`
`= 2 xx 10^(-3) m`.
`:. beta = ((5 xx 10^(-7)) (2.0))/(2 xx 10^(-3)) = 5 xx 10^(-4) m`
`= 0.05 cm`
Therefore, number of fringes ` =(P_(1) P_(2))/(beta) = 40`