question_answer

A thin film of soap solution \[({{\mu }_{s}}=1.4)\] lies on the top of a glass plate \[({{\mu }_{g}}=1.5)\]. When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths 420 nm and 630 nm. The minimum thickness of the soap solution are

A)  420 nm          

B)  500 nm

C)  450 nm           

D)  490 nm

Correct Answer: C

Solution :

For reflection at the air-soap solution interface, the phase difference is \[\pi \]. For reflection at the interface of soap solution to glass also, there will be a phase difference of\[\pi \]. \[\therefore \] The condition for the maximum intensity                 \[=2\mu t=n\lambda \] For n,    \[n{{\lambda }_{1}}=(n-1){{\lambda }_{2}}\]                 \[n\times 420=(n-1)\,\,630\] \[\therefore \] \[n\,(630-420)=630\] \[\Rightarrow \] \[n\,(210)=630\] \[\Rightarrow \] \[n=\frac{630}{210}\] \[\Rightarrow \] \[n=3\] This is the maximum order, where they coincide                 \[2\times 1.4\times t=3\times 420\] \[\Rightarrow \] \[t=\frac{3\times 420}{2\times 1.40}=450\,\,nm\]