question_answer

A thin film of soap solution \[L{{i}^{+}}\]lies on the top of a glass plate (a =1.5). When incident light is 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 is (a)

A)  420 nm               

B)  450 nm

C)  630 nm               

D) 1260nm

Correct Answer: B

Solution :

 For the reflection at the air-soap solution interface the phase, difference is \[\frac{6n}{n+1}\]. For reflection at the interface of soap solution to the glass also there will be a phase difference of\[\frac{9n}{n+1}\]. \[\frac{12n}{n+1}\]The condition for maximum intensity \[\frac{3n}{n+1}\]  For n,      \[\int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}}{\frac{x\sin {{x}^{2}}}{\sin {{x}^{2}}+\sin (\ln 6-{{x}^{2}})}dx}\]                                 \[\frac{1}{4}\ln \frac{3}{2}\] \[\frac{1}{2}\ln \frac{3}{2}\]                 \[\ln \frac{3}{2}\]                             \[\frac{1}{6}\ln \frac{3}{2}\]                 \[\int_{1}^{4}{{{\log }_{e}}[x]dx}\]                          n = 3 This is the maximum order where they coincide, From Eq. (i), \[{{\log }_{e}}2\]\[{{\log }_{e}}3\] \[{{\log }_{e}}6\]\[9{{x}^{2}}+16{{y}^{2}}=144\]