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In a YDSE two him transparent sheet are ...

In a YDSE two him transparent sheet are used in front of the slits `S_(1)`&`S_(2)` having `mu_(1)=1.6`and `mu_(2)=1.4` respectively.If both sheets have thickness`t` ,the central maximum is observed at a distance of `5mm`from center`O`.Now the sheets are replaced by another two sheets each of refractive index `(mu_(1)-mu_(2))/(2)` but having thickness `t_(1)&t_(2)=(t_(1)-t_(2))/(2)`.Now central maximum is observed at distance of `8mm` from center`O` on the same side as before.Find the thickness `t_(1)` (in`mu`m) [Given:d=1mm,D=1m]

Text Solution

Verified by Experts

The correct Answer is:
`33`
Path difference =`(S_(2)P-t+mu_(2)t)-(S_(1)P-t+mu_(1)t)`
`=(S_(2)P-S_(1)P)+(mu_(2)-mu_(1))t`
`=(x_(1)d)/(D)+(mu_(2)-mu_(1))t….(1)`
For central maximum at `P`, path difference=`0`
`(x_(1)d)/(mu_(2)-mu_(1))t=0 rArr (x_(1)d)/(D)=(mu_(1)-mu_(2))t`
`5(d)/(D)=(1.6-1.4)t rArr 0.2t=(5xx10^(-3)xx1xx10^(-3))/(1)`
`t=25mu m`
since `(mu_(1)+mu_(2))/(2)=1.5`
Now path difference =`{S_(2)P-t_(2)+(mu_(1)+mu_(2))/(2)t_(2)}-{S_(1)P-T_(1)+(mu_(1)+mu_(2))/(2)t_(1)}`
`=S_(2)P-S_(1)P+(t_(1)-t_(2))+(mu_(1)-mu_(2))/(2) (t_(2)-t_(1))`
`=(x_2d)/D+(t_1-t_2)+(mu_1+mu_2)/2(t_2-t_1)`
For central maximum path difference=`0`
`(x_(2)d)/(5)+(t_(1)-t_(2))+(mu_(1)+mu_(2))/(2)(t_(2)-t_(1))=0 rArr (x_(2)d)/(5)=t_(1)-t_(2)((mu_(1)+mu_(2))/(2)-1)`
`(8xx10^(-3)xx11xx10^(-3))/(1) =(t_(1)-t_(2))(1.5-1)`
`t_(1)-t_(2)=16xx10^(-6)`
`t_(1)-t_(2)=16mum...(2)`
Given that `(t_(1)+t_(2))/(2)=t rArr t_(1)+t_(2)=2t`
`t_(1)+t_(2)=50mum...(3)`
from(2)&(3)
`2t_(1)=66mum`
`t_(1)=33mum`
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