In a YDSE a parallel beam of light of wavelength 6000Å is incident on slits at angle of incidence `30^(@)`. A and B are two thin transparent films each of refractive index 1.5. thickness of A is `20.4mum`. Light coming through A and B have intensities I and 4I resepctively on the screen. Intensity at point O which is symmetric relative to the slits is 3I. The central maxima is above O.
(i). what is the maximum thickness of B to do so.
Assuming thickness of B to be that found in part (i)
Answer the following parts.
(ii). Find fringe width, maximum intensity and minimum intensity on screen.
(iii). Distance of nearest minima from O
(iv) intensity at 5 cm on either side of O.

Answer

(i). `t_(B)=120mum`
(ii). `beta=6mm,I_(max)=9I,I_(min)=I`
(iii). `(beta)/(6)=1mm`
(iv). `9I,3I`

Explanation

Path difference `=dcos60^(@)`
As give at 0 the intensity `=3I`
`3I+I+4I+2sqrt(I)sqrt(4I)cosphi`
`phi=(2pi)/(3)`
`(2pi)/(lamda)[(d)/(2)+(1.5-1)xx20.4xx10^(-6)-(1)/(2)txx10^(-6)]=(2pi)/(3)`
`implies0.1xx10^(-3)+(20.4-t)10^(-6)`
`=2//3xx6000xx10^(-10)=4xx10^(-7)`
`impliest=20.4-0.4+100=120mum`