Calculate the thickness of a quartz quarter- wave plate for sodium light `(lambda=5893Å)` given that the index of refraction of quartz for ordinary and extraordinary rays are 1.5442 and 1.5533 respectively.
Correct Answer - 1.45
The path difference is `2mu t`.
Now for destructive interface it can be
`2mut=(lambda)/(2)` or `(3lambda)/(2)` or `(5lambda)/(2)` and so on………
`mu=(lambda)/(4 t),(3lambda)/(4 t),(5lambda)/(4 t)……..`
`=(580xx10^(-9))/(4xx0.3xx10^(-6)) ,...
We have `[(lambda^(2)-2lambda+1,lambda-2),(1-lambda^(2)+3lambda,1-lambda^(2))]=Alambda^(2)+Blambda+C`
Putting `lambda=0`, we get
`C=[(1,-2),(1,1)]`
Putting `lambda=1`, we get
`A+B+C=[(0, -1),(3,0)]` ...(1)
Putting `lambda=-1`, we get
`A-B+C=[(4,-3),(-3,0)]`
Subtracting (2) from (1), we get
`2B[(0, -1),(3,0)]-[(4,-3),(-3,0)]=[(-4,2),(6,0)]`
`:. B=[(-2,...
Correct Answer - D
`(d)` `|{:(lambda-1,3lamda+1,2lambda),(lambda-1,4lambda-2,lambda+3),(2,3lambda+1,3(lambda-1)):}|=0`
`implieslambda=0` or `3`
If `lambda=0`, the equations become
`-x+y=0`,
`-x-2y+3z=0` and
`2x+y-3z=0`,
`:. (x)/(6-3)=(y)/(6-3)=(z)/(-1+4)`