A) 3:2
B) 1:25
C) 9:4
D) 1:5
Correct Answer: C
Solution :
We know that, \[\frac{{{\text{I}}_{\max }}}{{{\operatorname{I}}_{\min }}}=\frac{{{\left( \sqrt{\frac{{{\omega }_{1}}}{{{\omega }_{2}}}}+1 \right)}^{2}}}{{{\left( \sqrt{\frac{{{\omega }_{1}}}{{{\omega }_{2}}}}-1 \right)}^{2}}}\] \[{{\operatorname{I}}_{max}}\]and\[{{\operatorname{I}}_{\min }}\] are maximum and minimum intensity \[{{\omega }_{1}}\]and\[{{\omega }_{2}}\]are widths of two slits \[\therefore \]\[\frac{{{\operatorname{I}}_{max}}}{{{\operatorname{I}}_{min}}}=\frac{{{\left( \sqrt{\frac{1}{25}}+1 \right)}^{2}}}{{{\left( \frac{\sqrt{1}}{25}-1 \right)}^{2}}}\left( \frac{{{\omega }_{1}}}{{{\omega }_{2}}}=\frac{1}{25}\,\,\text{given} \right)\] On solving we get,\[\frac{{{\operatorname{I}}_{max}}}{{{\operatorname{I}}_{min}}}=\frac{\frac{36}{25}}{\frac{16}{25}}=\frac{9}{4}=9:4\]
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