A) \[\frac{9}{4}\]
B) \[\frac{121}{49}\]
C) \[\frac{49}{121}\]
D) \[\frac{4}{9}\]
Correct Answer: A
Solution :
| Given, In YDSE experiment, having two slits of width in the ratio of 1 : 25 |
| So, ratio of intensity, |
| \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{1}{25}\Rightarrow \frac{{{I}_{2}}}{{{I}_{1}}}=\frac{25}{1}\] |
| \[\therefore \] \[\frac{{{I}_{\max }}}{{{I}_{\max }}}=\frac{{{(\sqrt{{{I}_{2}}}+\sqrt{{{I}_{1}}})}^{2}}}{{{(\sqrt{{{I}_{2}}}-\sqrt{{{I}_{1}}})}^{2}}}={{\left[ \frac{\sqrt{\frac{{{I}_{2}}}{{{I}_{1}}}}+1}{\sqrt{\frac{{{I}_{2}}}{{{I}_{1}}}}-1} \right]}^{2}}\] |
| \[\Rightarrow \] \[{{\left[ \frac{5+1}{5-1} \right]}^{2}}={{\left( \frac{6}{4} \right)}^{2}}=\frac{36}{16}=\frac{9}{4}\] |
| Thus, \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{9}{4}\] |
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