Question

Two wavelengths of sodium light 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2×10
−6
m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.

Answer 1

Answer:

Two wavelengths of sodium light 590 nm, 596 nm are used, in turn, to study the diffraction taking place at a single slit of aperture 2 × 10-6 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the position of first maximum of the diffraction pattern obtained in the two cases.

Answer 2

Answer:

`lambda_(1) = 590 nm = 590 xx 10^(-9)m`  

`lambda_(2) = 596 nm = 596 xx 10^(-9) m`  

`a = 2 xx 10^(-6) m, D = 1.5 m`  

For first secondary maxima, ,brgt `sin theta = (2n + 1)(lambda)/(2a) = (3)/(2)(lambda)/(a) = (x)/(D)`,  

when `theta` is small.  

`:. x = (3 lambda D)/(2 a)`  

For two wavelengths,  

`x_(1) = (3lambda_(1) D)/(2 a) , x_(2) = (3lambda_(2) D)/(2 a)`  

Separation between the positions of first maxima  

`x_(2) - x_(1) = (3 (lambda_(2) - lambda_(1))D)/(2a)`  

`= (3 xx (596 - 590) xx 10^(-9) xx 1.5)/(2 xx 2 xx 10^(-6))`  

`= 4.5 xx 10^(-3) m = 4.5 mm`