In the experiment of interfernece, p is the number of bright bands for a light of wavelength `lamda_(1)`. If the source of light is replaced by `lamda_(2)` then the number of bright bands will be

Options

A. `(plamda_(2))/(lamda_(1))`
B. `(plamda_(1))/(lamda_(2))`
C. `plamda_(1)`
D. `plamda_(2)`

Answer

B

Explanation

We assum that the width of the region, where the interference fringes are formed is the samme in both the cases. Then if `X_(1) and X_(2)` are the fringe widths corresponding to `lamda_(1) and lamda_(2)` and `n_(1) and n_(2)` are the number of bright bands for `lamda_(1) and lamda_(2)` respectiively, then
`n_(1)X_(1)=n_(2)X_(2)` but `X=(lamdaD)/(d)`
`therefore n_(1)(lamda_(1)D)/(d)=n_(2)lamda_(2)(D)/(d)" [D and d are constant]"`
`therefore n_(1)lamda_(1)=n_(2)lamda_(2)`
`therefore n_(2)=(n_(1)lamda_(2))/(lamda_(2))` ltbgt In this case `n_(!)=p` (given)
`therefore n_(2)=(plamda_(1))/(lamda_(2))`