InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
In YDSE, bichromatic light of wavelengths 400 nm and 560 nm are used. The distance between the slits is 0.1 mm and the distance between the plane of the slits and the screen is 1m. The minimum distance between two successive regions of complete darkness isA. `4mm`B. `5.6mm`C. `14mm`D. `28mm` |
|
Answer» Correct Answer - D Let nth minima of 400 nm coincides with mth minima of 560 nm, then `(2n - 1) ((400)/(2)) = (2m - 1) ((560)/(2))` or `(2n-1)/(2m-1) = (7)/(5) = (14)/(10) =` … i.e., `4^(th)` minima of 400 nm coincides with `3^(rd)` minima of 560 nm. Location of this minima is, `Y_(1) = ((2xx4-1)(100)(400xx10^(-6)))/(2xx0.4) = 14 nm` Next `11^(th)` minima of 400 nm coincides with `8^(th)` minima of 560 nm, then `(2n-1) ((400)/(2)) = (2m - 1) ((560)/(2))` or `(2n-1)/(2m-1) = (7)/(5) = (14)/(10)= ` ... i.e., `11^(th)` minima of 400 nm coincides with `8^(th)` minima of 560 nm. Location of this minima is, `+Y_(2)` = `((2xx11-1)(100)(400xx10^(-6)))/(2xx0.1) = 42mm` Required distance `= Y_(2)-Y_(1) = 28 mm` Hence, the correct optical is (D). |
|