What is the highest spectral order that can be seen if a grating with 6500 slits per cm is illuminated with 633-nm laser light? Assume normal incidence.

Makenna Booker
2022-07-17
Answered

What is the highest spectral order that can be seen if a grating with 6500 slits per cm is illuminated with 633-nm laser light? Assume normal incidence.

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Aryan Novak

Answered 2022-07-18
Author has **6** answers

Given values:

Number of slits per centimeter, $N=6500\text{slits/cm}$

Wavelength, $\lambda =633\text{nm}$

Angle of diffraction, $\theta ={90}^{\circ}$

The phenomenon of bending of the light waves around the cornerns of obstacle is called diffraction.

The condition for diffraction maximum is given by,

$n\lambda =d\mathrm{sin}\theta $

Where,

n= Order of diffraction

$\lambda =$ Wavelength of light

d= Separation between the slits

$\theta =$ Angle of diffraction

The number of slits per centimeter is equal to the separation between the slits.

Number of slits per cm, $N=\frac{1}{d}$

Substituting equation

$n\lambda =\frac{\mathrm{sin}\theta}{N}\phantom{\rule{0ex}{0ex}}n=\frac{\mathrm{sin}\theta}{\lambda N}$

Substituting number of slits per centimeter, wavelength and angle of diffraction in equation

$n=\frac{\mathrm{sin}{90}^{\circ}}{6500\times {10}^{2}\text{slits/m}\times 633\times {10}^{-9}\text{m}}\phantom{\rule{0ex}{0ex}}=2.43\phantom{\rule{0ex}{0ex}}\approx 2$

Number of slits per centimeter, $N=6500\text{slits/cm}$

Wavelength, $\lambda =633\text{nm}$

Angle of diffraction, $\theta ={90}^{\circ}$

The phenomenon of bending of the light waves around the cornerns of obstacle is called diffraction.

The condition for diffraction maximum is given by,

$n\lambda =d\mathrm{sin}\theta $

Where,

n= Order of diffraction

$\lambda =$ Wavelength of light

d= Separation between the slits

$\theta =$ Angle of diffraction

The number of slits per centimeter is equal to the separation between the slits.

Number of slits per cm, $N=\frac{1}{d}$

Substituting equation

$n\lambda =\frac{\mathrm{sin}\theta}{N}\phantom{\rule{0ex}{0ex}}n=\frac{\mathrm{sin}\theta}{\lambda N}$

Substituting number of slits per centimeter, wavelength and angle of diffraction in equation

$n=\frac{\mathrm{sin}{90}^{\circ}}{6500\times {10}^{2}\text{slits/m}\times 633\times {10}^{-9}\text{m}}\phantom{\rule{0ex}{0ex}}=2.43\phantom{\rule{0ex}{0ex}}\approx 2$

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