Light with wavelength `lambda = 0.60 mu m` falls normally on a diffraction grating inscribed on a plane surfcae of a plano-curvex cylinderical glass lens with curvature radius `R = 20 cm`. The period of the grating is equal to `d = 6.0 mu m`. Find the distance between the principle maxima of first order located symmetrically in the focal plane of that lens.

Explanation

For the lens
`(1)/(f) = (n-1) ((1)/(R )-(1)/(oo))` or `f =(R )/(n - 1)`
For the grating
`d sin theta_(1) =lambda` or `sintheta_(1) =(lambda)/(d)`
`cosec theta_(1)=(d)/(lambda), cot theta_(1) = sqrt(((d)/(lambda))^(2) - 1)`
`tan theta_(1) = (1)/(sqrt(((d)/(lambda))^(2) - 1))`
Hence the distance between the two symmetrically placed first order maxima
`=2f tan theta_(1) = (2R)/((n -1)sqrt(((d)/(lambda))^(2) - 1))`
On putting `R = 20, n = 1.5, d=6.0mu m`
`lambda=0.60mu m` we get `8.04 cm`.