A commercial diffraction grating has 500 lines per mm. When a student shines a 530...

Step 1
The diffraction grating condition is,
${\displaystyle d\sin \theta =m\lambda }$
${\displaystyle m=\frac{d\sin \theta }{\lambda }}$
${\displaystyle =\frac{\left(\frac{1}{500\frac{l\in es}{mm}}\right)\sin 90}{530\times {10}^{-9}m}}$
${\displaystyle =\frac{\left(\frac{{10}^{-3}m}{500}\right)\sin 90}{530\times {10}^{-9}m}}$
${\displaystyle =3.77}$
${\displaystyle =3}$
Since for the maximum diffraction the angle is 90 degrees
So, the number of bright spots is,
${\displaystyle n=2m+1=2\left(3\right)+1=7}$

Step 1 As per the given data, $N=500$ and $\lambda =530nm$ Therefore, $d=\frac{1}{N}$ $=\frac{1mm}{500}$ $=2\times {10}^{-6}m$ $\therefore d=2\times {10}^{-6}m$ Step 2 In diffraction method, 1) $d\sin \theta =m\lambda$ Where, d is the width of the slit; m is the number of bright fringes on one side of central maxima; $\lambda$ is the wavelength of the light used; And $\theta$ is the angle made by the light with bright fringe To find maximum bright fringes, $\theta ={90}^{\circ }$ Step 3 Substitute all the given values in equation (1), $(2\times {10}^{-6})(\sin {90}^{\circ })=m(530\times {10}^{-9})$ $m=\frac{(2\times {10}^{-6})}{(530\times {10}^{-9})}$ $=\frac{2}{530}\times {10}^3$ $=3.7$ $\therefore m=3$ Since it is always a whole number This means, the number of bright fringes above the central maxima are 3 and also below the central maxima are 3 Therefore, $M=2m+1$ $=7$