A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
: The condition for an interference maximum in the light reflected from an soap film of thickness t is \[2\mu t\left( n+\frac{1}{2} \right)\] where \[n=0,1,2,.....\] and that for interference minimum is \[2\mu t=n\lambda \]where \[n=1,2,3.......\] Here, \[\mu =1.5\]. Now there is a maximum for \[\lambda =600nm\]and there is minimum for \[\lambda =450nm\] \[\therefore \] \[2\times 1.5\times t=\left( n+\frac{1}{2} \right)600\] ??(i) \[2\times 1.5\times t=(n+1)450\] ??(ii) In view of Eq. (i), we have taken the integer \[(n+1)\] rather than n in Eq. (ii) Comparing (i) and (ii), we get \[\left( n+\frac{1}{2} \right)600=(n+1)450\] \[\therefore \] \[n=1\] Substituting n = 1, in Eq. (i), we get \[2\times 1.5\times t=\frac{3}{2}\times 600nm\] \[t=\frac{3\times 600\times {{10}^{-9}}}{2\times 2\times 1.5}=3\times {{10}^{-7}}m\]
You need to login to perform this action.
You will be redirected in
3 sec
