Question

In Youngs double slit experiment, the intensity of light coming from one of the slits is doubled the intensity from the other slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed is

• A. 34
• B. 40
• C. 25
• D. 38

Answer 1

Answer: (A)

$\frac{I_{\max }}{I_{\min }}=\frac{(\sqrt{2I}+\sqrt{I}{)}^2}{(\sqrt{2I}-\sqrt{I}{)}^2}={\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)}^2$
$={\left[\frac{(\sqrt{2}+1)(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}\right]}^2$
$={\left(\frac{2+1+2\sqrt{2}}{2-1}\right)}^2$
$=\frac{9+8+12\sqrt{2}}{1}$
$\approx 34$