Solution :
Given :
The angle of the first maximum with the center is given by :
[tex]$a=\tan^{-1}\left(\frac{0.488}{1.88}\right)$[/tex]
= 14.5°
The grating distance can be calculated as :
[tex]$d=\frac{1 \ cm}{5308 \text{ slits}}$[/tex]
= [tex]$1.88 \times 10^{-4} \ m$[/tex]
When the principal maxima yields at y = 0.488 m and the length from the wall 1.88 m. Thus the equation of the wavelength is :
[tex]$\lambda = g \times \frac{\sin a}{n}$[/tex] , where n = 1
[tex]$=1.88 \times 10^{-4} \times \sin (14.5)$[/tex]
[tex]$=4.70 \times 10^{-5} \ m$[/tex]
