A) 2
B) 3
C) 4
D) 5
Correct Answer: B
Solution :
From \[\Delta {{S}_{1}}{{S}_{2}}D,\] \[{{({{S}_{1}}D)}^{2}}={{({{S}_{1}}{{S}_{2}})}^{2}}+{{({{S}_{2}}D)}^{2}}\] \[{{({{S}_{1}}P+PD)}^{2}}={{({{S}_{1}}{{S}_{2}})}^{2}}+{{({{S}_{2}}D)}^{2}}\] Here \[{{S}_{1}}P\] is the path difference \[=n\lambda \] for maximum intensity. \[\therefore {{(n\lambda +{{x}_{n}})}^{2}}={{(4\lambda )}^{2}}+{{({{x}_{n}})}^{2}}\] or \[{{x}_{n}}=\frac{16{{\lambda }^{2}}-{{n}^{2}}{{\lambda }^{2}}}{2n\lambda }\] Then \[{{x}_{1}}=\frac{16{{\lambda }^{2}}-{{\lambda }^{2}}}{2\lambda }=7.5\,\lambda \] \[{{x}_{2}}=\frac{16{{\lambda }^{2}}-4{{\lambda }^{2}}}{4\lambda }=3\lambda \] \[{{x}_{3}}=\frac{16{{\lambda }^{2}}-9{{\lambda }^{2}}}{6\lambda }=\frac{7}{6}\lambda \] \[{{x}_{4}}=0\]. \[\therefore \]Number of points for maxima becomes 3.You need to login to perform this action.
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