Practicing Success
Practicing Success
CUET Preparation Today
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a b c d |
b |
$\frac{I_1}{I_2} = \beta$ $ I_{max} =( \sqrt I_1 + \sqrt I_2)^2 $ $I_{min} =( \sqrt I_1 - \sqrt I_2)^2 $ $\frac{I_{max} }{I_{min}} = \frac{( \sqrt I_1 + \sqrt I_2)^2 }{( \sqrt I_1 - \sqrt I_2)^2}= \frac{2\sqrt \beta}{1+\beta}$ |
