Joseph Krupa

2022-01-17

A commercial diffraction grating has 500 lines per mm. When a student shines a 530 nm laser through this grating, On the screen behind the grating, how many bright spots are visible?

Papilys3q

Expert

Step 1
The condition for the diffraction grating is,
$d\mathrm{sin}\theta =m\lambda$
The angle is 90 degrees for bright fringes with the highest order. Any angle greater than 90 degrees will result in the diffraction going behind itself, which is not possible.
$\left(\frac{1}{500}×{10}^{-3}\right)\mathrm{sin}90=m\left(530×{10}^{-9}m\right)$
$\left(\frac{1}{500}×{10}^{-3}\right)\mathrm{sin}90$
$=3.77$
Choose the integer value of m
So, the number of bright spots is,
$n=2m+1=2×3+1=7$
Hence, the number of bright spots is 7.

Piosellisf

Expert

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

alenahelenash

Expert

Step 1 As per the given data, $N=500$ and $\lambda =530nm$ Therefore, $d=\frac{1}{N}$ $=\frac{1mm}{500}$ $=2×{10}^{-6}m$ $\therefore d=2×{10}^{-6}m$ Step 2 In diffraction method, 1) $d\mathrm{sin}\theta =m\lambda$ 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), $\left(2×{10}^{-6}\right)\left(\mathrm{sin}{90}^{\circ }\right)=m\left(530×{10}^{-9}\right)$ $m=\frac{\left(2×{10}^{-6}\right)}{\left(530×{10}^{-9}\right)}$ $=\frac{2}{530}×{10}^{3}$ $=3.7$ $\therefore m=3$ 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, $M=2m+1$ $=7$