Question

In young's double slit experiment, the ratio of maximum to minimum intensities of the fringe system is 4:1. The amplitudes of the coherent sources are in the ratio
A] 4:1 B] 3:1 C] 2:1 D] 1:1

Answer 1

Dear Student,

$$\begin{gathered} \frac{{\mathrm{I}}_{\max }}{{\mathrm{I}}_{\min }}=\frac{4}{1} \\ \mathrm{Also},\frac{{\mathrm{I}}_{\max }}{{\mathrm{I}}_{\min }}=\frac{{\left(\mathrm{a}+\mathrm{b}\right)}^2}{{\left(\mathrm{a}-\mathrm{b}\right)}^2} \\ \frac{4}{1}=\frac{{\left(\mathrm{a}+\mathrm{b}\right)}^2}{{\left(\mathrm{a}-\mathrm{b}\right)}^2} \\ \sqrt{\frac{4}{1}}=\frac{\left(\mathrm{a}+\mathrm{b}\right)}{\left(\mathrm{a}-\mathrm{b}\right)} \\ \frac{\left(\mathrm{a}+\mathrm{b}\right)}{\left(\mathrm{a}-\mathrm{b}\right)}=\frac{2}{1} \\ \mathrm{a}+\mathrm{b}=2\mathrm{a}-2\mathrm{b} \\ \mathrm{a}=3\mathrm{b} \\ \frac{\mathrm{a}}{\mathrm{b}}=\frac{3}{1} \\ \mathrm{where},\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{are}\mathrm{the}\mathrm{amplitudes}. \end{gathered}$$