Saved Bookmarks
| 1. |
In Young's double slit experiment the two slits 0.15mm apart are illuminated by monochromatic light of wavelength 450nm. The screen is 1.0m away from the slits. (a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maxima. (b) How will the fringe pattern change if the screen is moved away from the slits ? |
|
Answer» Solution :(a) Distance of `n^(th)` maxima from central maxima `x_(n)=(nlambdaD)/(d)`, where `, n=0`,`+-1`, `+2`…… Given : `n=2, d=0.15mm, lambda=450nm` and `D=1.0m` `x_(2)=(2xx450xx10^(-9)xx1.0)/(0.15xx10^(-3))=6XX10^(-3)m=6mm` (b) Distance of `n^(th)` minima from central maxima `y_(2)=((2n-1)lambdaD)/(2D)=((2xx2-1)450xx10^(-9))/(2xx0.5xx10^(-3))` `=4.5=10^(-3)m=4.5mm` When the screen is moved AWAY from the slits fringes become FARTHER apart. ( `:.` Fringe width `prop` distance of screen). |
|