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), \[{{\log }_{e}}2\]\[{{\log }_{e}}3\] \[{{\log }_{e}}6\]\[9{{x}^{2}}+16{{y}^{2}}=144\]You need to login to perform this action.
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