Emmanuel Pace

2022-07-19

The wavelengths of the visible spectrum range from about 400 nm (purple) to 700 nm (red). Find the angular width of the first-order visible spectrum created by a diffraction grating with 565 lines per millimeter when white light falls on the grating normally. Give the answer in degrees.

polishxcore5z

Beginner2022-07-20Added 14 answers

We have to calculate the angular width.

Here the wavelength of the visible spectrum are 400 nm to 700 nm and there are 565 lines per mm.

So the grating distance d=565 lines (mm)$=565\times {10}^{-3}\text{}m$

We represent that the maximum order integer value m at the observable angle $\theta ={90}^{\circ}$

Thus $d\mathrm{sin}\theta =m{\lambda}_{red}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}{\lambda}_{red}=700\text{}nm=7\times {10}^{-9}m$

So $m=\frac{d}{\lambda}\mathrm{sin}\theta $

If the anglar width be $d\theta $ then

${\lambda}_{red}-{\lambda}_{violet}=d\mathrm{sin}(d\theta )$

So $\mathrm{sin}(d\theta )=\frac{{\lambda}_{red}-{\lambda}_{violet}}{d}=\frac{7\times {10}^{-9}-4\times {10}^{-9}}{565\times {10}^{-3}}$

So $d\theta =3.03\times {10}^{-7}$

Thus the angular width is $(3.03\times {10}^{-7})$

Here the wavelength of the visible spectrum are 400 nm to 700 nm and there are 565 lines per mm.

So the grating distance d=565 lines (mm)$=565\times {10}^{-3}\text{}m$

We represent that the maximum order integer value m at the observable angle $\theta ={90}^{\circ}$

Thus $d\mathrm{sin}\theta =m{\lambda}_{red}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}{\lambda}_{red}=700\text{}nm=7\times {10}^{-9}m$

So $m=\frac{d}{\lambda}\mathrm{sin}\theta $

If the anglar width be $d\theta $ then

${\lambda}_{red}-{\lambda}_{violet}=d\mathrm{sin}(d\theta )$

So $\mathrm{sin}(d\theta )=\frac{{\lambda}_{red}-{\lambda}_{violet}}{d}=\frac{7\times {10}^{-9}-4\times {10}^{-9}}{565\times {10}^{-3}}$

So $d\theta =3.03\times {10}^{-7}$

Thus the angular width is $(3.03\times {10}^{-7})$

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