Mumbai university > FE > SEM 2 > Applied Physics II

Marks: 3M

Year: May 2013

Given:

Maximum angle of diffraction = $sin⁡Ѳ = 90^0$

Light of wavelength= $λ = 5896 A^o = 5896 × 10^{-10}m$

Maxima seen from diffraction grating=e = 5500 lines/cm = $\frac{10^{-2}}{5500} = 1.82 × 10^{-6}m$

Solution:

The expression for maximum order of diffraction is given by

$m = \frac{e sin⁡Ѳ}{λ}$

$m = \frac{1.82 × 10^{-6} × 1}{5896 × 10^{-10}}$

$m = 3.086$

Therefore maximum number of orders = 3, and a total of seven images of the source can be seen (three on each side of a central image).