JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 4)
In an interference experiment the ratio of amplitudes of coherent waves is $${{{a_1}} \over {{a_2}}} = {1 \over 3}$$ . The
ratio of maximum and minimum intensities of
fringes will be :
2
4
18
9
Explanation
$${{{I_{\max }}} \over {{I_{\min }}}} = {{{{\left( {{a_1} + {a_2}} \right)}^2}} \over {{{\left( {{a_1} - {a_2}} \right)}^2}}} = {{{{\left( {1 + 3} \right)}^2}} \over {{{\left( {1 - 3} \right)}^2}}} = {{16} \over 4}$$ = 4
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