Saved Bookmarks
| 1. |
Consider the situation shown in figure. The two slits S_1 and S_2 placed symmetrically around the central line are illuminated by a monochromatic light of wavelength lamda. The separation between the slits is d. The light transmitted by the slits falls on a screen Sigma_1 placedat a distance D from the slits. The slit S_3 is at the placedcentral line and the slit S_4, is at a distance z from S_3. Another screenSigma_2 is placed a further distance D away from 1,1. Find the ratio of the maximum to minimum intensity observed on Sigma_2 if z is equal to a. z=(lamdaD)/(2d) b. (lamdaD)/d c. (lamdaD)/(4d) |
|
Answer» minimum intensilty OCCURS, (dark fringe) `RARR Amplitude =0` At `S_3` path differenc e=0` rarr maximum intensity occurs. rarr Amplitude =2r So on `Sigma^2` screen, `l_(max)/l_(min)=((2r+0)^2)/((2r-0)^2)=1` ii. When z=(lamdaD)/d ` As `S_4` maximum intensity occurs `rarr Amplitude =sqrt2r` At `S_3` also maximum intensity occurs. `l_(max)/l_(min)=((2r+2r)^2)/((2r-2r)^2)=oo` iii When `z=(lamdaD)/(4d), at S_4` intensity =l_(max)/2` `rarr Amplitude=sqrt2r` `:. At S_3` Intensilty is minimum `rarr Amplitude =2r `l_(max)/l_(min)=((2r+sqrt(2r))^2)/((2r-sqrt(2r))^2)=34` |
|