Isaac Barry

2022-09-04

Diffraction gratings with 10,000 lines per centimeter are readily available. Suppose you have one, and you send a beam of white light through it to a screen 2.00 m away.Find the angles for the first-order diffraction of the shortest and longest wavelengths of visible light (380 and 760 nm).

Sanaa Hudson

Beginner2022-09-05Added 7 answers

Given information:

The number of lines on the grating, $N=10000\text{lines/cm}=10000\times \frac{\text{lines}}{{10}^{-2}\text{m}}={10}^{6}\text{lines/m}$

The distance between the screen and slits, $D=2.00\text{m}$

The shortest wavelength, ${\lambda}_{\text{min}}=380\text{nm}=380\times {10}^{-9}\text{m}$

The longest wavelength, ${\lambda}_{\text{max}}=760\text{nm}=760\times {10}^{-9}\text{m}$

To find

the angles corresponding to first-order diffraction from these wavelengths.

The grating equation is given by

$d\mathrm{sin}(\theta )=n\lambda \phantom{\rule{0ex}{0ex}}\mathrm{sin}(\theta )=\frac{n\lambda}{d}\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{sin}}^{-1}\frac{n\lambda}{d}$

where,

d=width of the slit

n=order of diffraction

$\lambda =$ wavelength of incident light

$\theta =$ angle of diffraction

The width of the slit is calculated as follows

$d=\frac{1}{N}\phantom{\rule{0ex}{0ex}}d=\frac{1}{{10}^{6}}\phantom{\rule{0ex}{0ex}}d={10}^{-6}\text{m}$

Apply the above formula for the shortest wavelength

${\theta}_{\text{min}}={\mathrm{sin}}^{-1}(\frac{1\times 380\times {10}^{-9}}{{10}^{-6}})\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}(0.38)\phantom{\rule{0ex}{0ex}}={22.33}^{\circ}$

Now, apply the above formula for the longest wavelength

${\theta}_{\text{max}}={\mathrm{sin}}^{-1}(\frac{1\times 760\times {10}^{-9}}{{10}^{-6}})\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}(0.76)\phantom{\rule{0ex}{0ex}}={49.46}^{\circ}$

Answer:

The angle corresponding to the shortest wavelength is, ${\theta}_{\text{min}}={22.33}^{\circ}$

The angle corresponding to the longest wavelength is, ${\theta}_{\text{max}}={49.46}^{\circ}$

The number of lines on the grating, $N=10000\text{lines/cm}=10000\times \frac{\text{lines}}{{10}^{-2}\text{m}}={10}^{6}\text{lines/m}$

The distance between the screen and slits, $D=2.00\text{m}$

The shortest wavelength, ${\lambda}_{\text{min}}=380\text{nm}=380\times {10}^{-9}\text{m}$

The longest wavelength, ${\lambda}_{\text{max}}=760\text{nm}=760\times {10}^{-9}\text{m}$

To find

the angles corresponding to first-order diffraction from these wavelengths.

The grating equation is given by

$d\mathrm{sin}(\theta )=n\lambda \phantom{\rule{0ex}{0ex}}\mathrm{sin}(\theta )=\frac{n\lambda}{d}\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{sin}}^{-1}\frac{n\lambda}{d}$

where,

d=width of the slit

n=order of diffraction

$\lambda =$ wavelength of incident light

$\theta =$ angle of diffraction

The width of the slit is calculated as follows

$d=\frac{1}{N}\phantom{\rule{0ex}{0ex}}d=\frac{1}{{10}^{6}}\phantom{\rule{0ex}{0ex}}d={10}^{-6}\text{m}$

Apply the above formula for the shortest wavelength

${\theta}_{\text{min}}={\mathrm{sin}}^{-1}(\frac{1\times 380\times {10}^{-9}}{{10}^{-6}})\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}(0.38)\phantom{\rule{0ex}{0ex}}={22.33}^{\circ}$

Now, apply the above formula for the longest wavelength

${\theta}_{\text{max}}={\mathrm{sin}}^{-1}(\frac{1\times 760\times {10}^{-9}}{{10}^{-6}})\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}(0.76)\phantom{\rule{0ex}{0ex}}={49.46}^{\circ}$

Answer:

The angle corresponding to the shortest wavelength is, ${\theta}_{\text{min}}={22.33}^{\circ}$

The angle corresponding to the longest wavelength is, ${\theta}_{\text{max}}={49.46}^{\circ}$

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