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 \[\text{42}0\text{ nm}\] and  \[\text{63}0\text{ nm}\]. The minimum thickness of the soap solution are

A) \[\text{42}0\text{ nm}\]              

B) \[~\text{5}00\text{ nm}\]

C) \[~\text{45}0\text{ nm}\]                           

D) \[\text{49}0\text{ nm}\]

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 \[2\times 1.4\times t=3\times 420\] \[\Rightarrow \]               \[\text{t =}\frac{\text{3 }\!\!\times\!\!\text{ 420}}{\text{2 }\!\!\times\!\!\text{ 1}\text{.40}}\text{= 450 nm}\]