Question

The intensity of the light coming from one of the slits in Young’s experiment is twice the intensity of the light coming from the other slit. What will be the approximate ratio of the intensities of the bright and dark fringes in the resulting interference pattern?

Answer 1

Given : I₁:I₂ = 2:1

If E₁₀ and E₂₀ are the amplitudes of the interfering waves, the ratio of the maximum intensity to the minimum intensity in the fringe system is

\(\frac{I_{max}}{i_{min}}=(\frac{E_{10}+E_{20}}{E_{10}-E_{20}})^2 =(\frac{r+1}{r-1})^2\)

where r = E₁₀/E₂₀.

∴ \(\frac{I_1}{I_2}=(\frac{E_{10}}{E_{20}})^2=r^2\)

∴ r =\(\sqrt{\frac{I_1}{I_2}}\)

∴ \(\frac{I_{max}}{i_{min}}=(\frac{\sqrt{2}+1}{\sqrt{2}-1})^2=(\frac{2.414}{0.414})^2\) = (5.83)² = 33.99 = 34

The ratio of the intensities of the bright and dark fringes in the resulting interference pattern is 34 : 1.