Saved Bookmarks
| 1. |
For a single-slit of width "a", the first minimum of the interference pattern of a monochromatic light of wavelength lamda occurs at an angle of (lamda)/(a). At the same angle of (lamda)/(a), we get a maximum for two narrow slits separated by a distance "a". explain. |
|
Answer» Solution :We know that for a single-slit of width .a. we obtain NTH ORDER minimum due to superposition of wavelets passing through the slit at an angle `THETA` such that path difference between extreme waves is given by `asintheta=nalamda` Hence, for first minima `n=1,asintheta_(1)=lamdaimpliestheta_(1)=sin^(-1)((lamda)/(a))`. and for small value of `theta_(1)`, we can say that `theta_(1)=(lamda)/(a)` On the other hand, for interference pattern due to superposition of light waves coming from two coherent sources separated by a distnace .a., we obtain a maximum if path difference between corresponding waves is given by : `asintheta=nalamda` and for first maximum (n=1), we have `asintheta_(1)=lamdaimplies sintheta_(1)=(lamda)/(a)` and for small value of `theta_(1)`, we can say that `sintheta_(1)=theta_(1)=(lamda)/(a)`. |
|