In an experiment of single slit diffraction pattern first minimum for red light coincide with first maximum of some other wavelength. If wavelength of red light is 6600 angstroms, then the wavelength of first maximum will be

Question

samigupta8 · Answer

@Michele_laino sir

samigupta8 · Answer

Sir I did this....

samigupta8 · Answer

We know that condition for secondary maximum in case of diffraction is path difference should be (2n-1)lambda/2...
N for minimum it is n lambda

michele_laino · Answer

I'm sorry I'm not good with physical optics

samigupta8 · Answer

And since question says that they both are first minimum and maximum so we put n=1 in both of them

samigupta8 · Answer

Oh...it"s alright sir....i will ask from somebody else...:)

michele_laino · Answer

ok! :)

samigupta8 · Answer

@Vincent-lyon.fr

samigupta8 · Answer

@mayankdevnani

vincent-lyon.fr · Answer

Simply draw a sketch of the pattern:
|dw:1456128127348:dw|

then write $\theta$ in terms of $\lambda_1$ and $\lambda_2$