A) \[{{I}_{1}}=0.531{{I}_{0}}\]
B) \[{{I}_{1}}=0.053{{I}_{0}}\]
C) \[{{I}_{1}}=53{{I}_{0}}\]
D) \[{{I}_{1}}=5.03{{I}_{0}}\]
Correct Answer: A
Solution :
\[{{I}_{0}}={{R}^{2}}=\frac{R_{2}^{2}}{4}\] Number of HPZ covered by the disc at \[b=25\,cm\] \[{{n}_{1}}{{b}_{1}}={{n}_{2}}{{b}_{2}}\] \[{{n}_{2}}=\frac{{{n}_{1}}{{b}_{1}}}{{{b}_{2}}}+\frac{1\times 1}{0.25}=4\] Hence the intensity at this point is \[I={{{R}'}^{2}}={{\left( \frac{{{R}_{5}}}{2} \right)}^{2}}={{\left( \frac{{{R}_{5}}}{{{R}_{4}}}\times \frac{{{R}_{4}}}{{{R}_{3}}}\times \frac{{{R}_{3}}}{{{R}_{2}}} \right)}^{2}}\times {{\left( \frac{{{R}_{2}}}{2} \right)}^{2}}\] or \[1={{[0.9]}^{6}}\] \[{{I}_{1}}=0.531\,\,{{I}_{0}}\] Hence the correct answer will be (a).
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