1.

Angular width of central maximum in the Fraunhoffer's diffraction pattern is measured.Slit is illuminated by the light of wavelength 6000 Å. If slit illuminated by light of another wavelength, angular width decreased by 30%. Wavelength of light used is :

Answer»

`3500 Å`
`4200 Å`
`4700 Å`
`6000 Å`

Solution :` d SIN theta = n lambda`
For first diffraction, n = 1
`therefore d sin theta = lambda`
or `d theta = lambda` (`because sin theta = theta` for small angle)
For ANGULAR width, `theta = (lambda)/(d)`
For full angular width,
`OMEGA = 2 theta = (2 lambda)/(d)`
Again `omenag. = (2 lambda.)/(d)`
`therefore` DIVIDING we get.
`(omega.)/(omega) = (lambda.)/(lambda) ` or `lambda. = (lambda. omega.)/(omega)`
`therefore lambda. = 4200 Å`.


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