Saved Bookmarks
| 1. |
In an arrangement of double slit arrangement fig. The slits are illuminated by light of wavelenth 600 mm. The distance of the first point on the screen from the centre maximum where intensity is 75% of central maximum is |
|
Answer» Solution :`I_("max")= I_(1)= 4a^(2)""` (at CENTRAL maximum) `I_(2)= 75% " of "I_(1)""` (at point P) `=(3I)/(4)= (3)/(4)xx4a^(2)= 3A^(2)` Resultant AMPLITUDE at .P. is `A= sqrt(3)a` If `phi` is phase DIFFERENCE `A^(2)= A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2) cos phi` `3Aa^(2)= A^(2)+A^(2)+2A^(2)cos phi : phi =(pi)/(3)` Then corresponding path difference `phi = (2pi x)/(lambda) implies x= (lambda)/(6) therefore` Path difference `x= y(d)/(D)` `y= 4.8xx 10^(-5)m`. |
|