In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern
Answer & explanation
Correct answer: option 1
In the interference pattern of Young's double slit experiment maximum and minimum intensity is given as
$I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$
$I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$
Here $I_1$ has been doubled and $I_2$ remains same. Therefore both $I_1$ and $I_2$ will increase.
Hence (A) is correct