In young’s double slit experiment. Assume intensity of each source is I0 and K1 is equal to difference of maximum and minimum intensity. Now intensity of one source is made $\frac{I_0}{4}$ and K2 is again difference of maximum and minimum intensity. Then $\frac{K_1}{K_2} = $
Answer & explanation
Correct answer: option 4
$\mathrm{K}_1=4 \mathrm{I}_0-0$
In second case
$\mathrm{I}_{\max }=\mathrm{I}_0\left(\frac{1}{2}+1\right)^2=\frac{9}{4} \mathrm{I}_0 ; \mathrm{I}_{\min }=\mathrm{I}_0\left(1-\frac{1}{2}\right)^2=\frac{\mathrm{I}_0}{4}$
$\mathrm{~K}_2=\frac{9 \mathrm{I}_0}{4}-\frac{\mathrm{I}_0}{4}=2 \mathrm{I}_0 ; \frac{\mathrm{K}_1}{\mathrm{~K}_2}=2$