Question

The intensity of the light coming from one of the slits in a Young's double slit experiment is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed.

Answer 1

Let the intensity of light coming from one slit $I_1=I$

Intensity of light coming from other slit $I_2=2I$

Maximum intensity $I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2$

Or $I_{max}=(\sqrt{I}+\sqrt{2I}{)}^2=(\sqrt{2}+1{)}^{2}I$

Minimum intensity $I_{min}=(\sqrt{I_2}-\sqrt{I_1}{)}^2$

Or $I_{min}=(\sqrt{2I}-\sqrt{I}{)}^2=(\sqrt{2}-1{)}^{2}I$

Ratio $\frac{I_{max}}{I_{min}}=\frac{(\sqrt{2}+1{)}^2}{(\sqrt{2}-1{)}^2}$