Light with wavelength `lambda = 0.50mu m` falls on a slit of width `b = 10mum` at an angle `theta_(0) = 30^(@)` to its normal. Find the angular position of the first minima located on both sides of the central Fraunhofer maximum.

Explanation

The relation `b sin theta = k lambda`
for minima (which light is incident normally on the slit) has a simple interpertation `:b sin theta` is the path difference between exterme wave normals emitted at angle `theta`

When light is incident at an angle `theta_(0)` the path difference is
`b(sin theta - sin theta_(0))`
and the condition of minima is
`b(sin theta - sin theta_(0)) = k lambda`
For the first minima
`b (sin theta - sin theta_(0)) = +- lambda` or `sin theta = sin theta_(0) +-(lambda)/(b)`
Putting in numbers `theta_(0) = 30^(@), lambda = 0.50mu m b = 10mu m`
`sin theta = (1)/(2) +- (1)/(20) = 0.55` or `0.45`
`theta_(+1) = 33^(@) - 20'` and `theta_(-1) = 26^(@) 44'`