Saved Bookmarks
| 1. |
The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is |
|
Answer» infinite `d SIN theta=(lambda)/(d)=(nlambda)/(2lambda)"" [ :. D=2lambda]` `:. sin teta=(n)/(2)` but RANGE of `sin theta` is(-1,1) `:. Sin theta le1` `:. n le2` `:.` Possible values of `n=-2,1,0,1,2` Hence, a central maximum on the screen with two sides of the FIRST order and two of the second order respectively gives a total of five bright fringes. |
|