The width of one of the two slits in a Young’s double slit experiment is three times the other slit. If the amplitude of light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is x:4 where x is _ _ _ _ .

Solution

Given, $\frac {A_1}{A_2} = \frac {3}{1}$

Now, $\frac {I_1}{I_2} = \frac {A_1 ^2 }{A_2 ^2} = (\frac {3}{1})^2 = \frac {9}{1}$

$\frac {I_{min}}{I_{max}} = \frac {(\sqrt I_1 – \sqrt I_2 )^2 }{(\sqrt I_1 + \sqrt I_2 )^2} = \frac {(3-1)^2}{(3+1)^2}=1:4$

Answer: 1.00