In a grating spectrum, which spectral line in fourth order will overlap with third order line of $5461{A}^{0}$

vangstosiis
2022-07-17
Answered

In a grating spectrum, which spectral line in fourth order will overlap with third order line of $5461{A}^{0}$

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Kaiden Weeks

Answered 2022-07-18
Author has **14** answers

Given

The wavelength of a spectral line in third-order is ${\lambda}_{3}=5461{A}^{0}$

The integer of the third order is ${n}_{3}=3.$

The integer of the fourth-order is ${n}_{4}=4.$

If the grating spectrum is the same for both the orders then,

$n\lambda =\text{constant}\phantom{\rule{0ex}{0ex}}{n}_{3}{\lambda}_{3}={n}_{4}{\lambda}_{4}$

Here, ${\lambda}_{4}$ is the wavelength of a spectral line in fourth-order.

Substitute the known values.

$3\times 5461{A}^{0}=4\times {\lambda}_{4}\phantom{\rule{0ex}{0ex}}{\lambda}_{4}=\frac{3\times 5461{A}^{0}}{4}\phantom{\rule{0ex}{0ex}}=4095.75{A}^{0}$

Thus, in the grating spectrum, the wavelength of the spectral line of fourth-order is $4095.75{A}^{0}$

The wavelength of a spectral line in third-order is ${\lambda}_{3}=5461{A}^{0}$

The integer of the third order is ${n}_{3}=3.$

The integer of the fourth-order is ${n}_{4}=4.$

If the grating spectrum is the same for both the orders then,

$n\lambda =\text{constant}\phantom{\rule{0ex}{0ex}}{n}_{3}{\lambda}_{3}={n}_{4}{\lambda}_{4}$

Here, ${\lambda}_{4}$ is the wavelength of a spectral line in fourth-order.

Substitute the known values.

$3\times 5461{A}^{0}=4\times {\lambda}_{4}\phantom{\rule{0ex}{0ex}}{\lambda}_{4}=\frac{3\times 5461{A}^{0}}{4}\phantom{\rule{0ex}{0ex}}=4095.75{A}^{0}$

Thus, in the grating spectrum, the wavelength of the spectral line of fourth-order is $4095.75{A}^{0}$

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