Question

The human eye can readily detect wavelengths from about 400 nm to 700 nm. If white light illuminates a diffraction grating having 750 lines/mm, over what range of angles does the visible m = 1 spectrum extend?

 

Answer 1

Concepts and reason

The concept required to solve the question is diffraction through the grating. First find the angle does the visible \(m=1\) spectrum at \(400 \mathrm{nm}\) using the condition for bright fringes in a diffraction grating. Later calculate the angle does the visible \(m=1\) spectrum at \(700 \mathrm{nm}\) using the condition for bright fringes in a diffraction grating. Finally calculate the range of angles does the visible \(m=1\) spectrum extend.

Fundamentals

Condition for the bright fringes in a diffraction grating is, \(d \sin \theta=m \lambda\)

Here, \(\lambda\) is the wavelength, \(\theta\) is the angle, \(\mathrm{m}\) is the order number of diffraction pattern and \(d\) is the distance on a grating.

 

Expression for the bright fringes in a diffraction grating is as follows:

\(d \sin \theta=m \lambda\)

Rearrange the above equation for \(\theta\).

\(\sin \theta=\frac{m \lambda}{d}\)

\(\theta=\sin ^{-1}\left(\frac{m \lambda}{d}\right)\)

Substitute 1 for \(\mathrm{m}, 400 \mathrm{nm}\) for \(\lambda\) and \(\left(\frac{1}{750}\right) \mathrm{mm}\) for \(\mathrm{d}\) in the above equation.

\(\theta=\sin ^{-1}\left(\frac{m \lambda}{d}\right)\)

\(=\sin ^{-1}\left(\frac{(1)\left(400 \times 10^{-9} \mathrm{~m}\right)}{(750) \mathrm{mm}}\right)\)

\(=\sin ^{-1}\left(\frac{(1)\left(400 \times 10^{-9} \mathrm{~m}\right)}{\left(1.33 \times 10^{-6} \mathrm{~m}\right)}\right)\)

\(=17.5^{\circ}\)

since, the angles of the diffracted modes are related to both wavelength and grating through the grating equation. And the resultant grating equation will only predicts the directions of the modes. Here, if the incident radiation contains several wavelengths, and each wavelength deviates along a specific line.

 

 

Expression for the bright fringes in a diffraction grating is as follows:

\(d \sin \theta=m \lambda\)

Rearrange the above equation for \(\theta\).

\(\sin \theta=\frac{m \lambda}{d}\)

\(\theta=\sin ^{-1}\left(\frac{m \lambda}{d}\right)\)

Substitute 1 for \(\mathrm{m}, 700 \mathrm{nm}\) for \(\lambda\) and \(\left(\frac{1}{750}\right) \mathrm{mm}\) for \(\mathrm{d}\) in the above equation.

\(\theta=\sin ^{-1}\left(\frac{m \lambda}{d}\right)\)

\(=\sin ^{-1}\left(\frac{(1)\left(700 \times 10^{-9} \mathrm{~m}\right)}{(750) \mathrm{mm}}\right)\)

\(=\sin ^{-1}\left(\frac{(1)\left(700 \times 10^{-9} \mathrm{~m}\right)}{\left(1.33 \times 10^{-6} \mathrm{~m}\right)}\right)\)

\(=31.75^{\circ}\)

The range of angles the visible \(m=1\) spectrum extend is from \(17.5^{\circ}\) to \(31.75^{\circ}\)

Here, the parallel lines on the diffraction grating are used to diffract light and generally used for both separating and spreading as different colors or wavelengths. And the spectrum is defined as the array of light, which is arranged by the order of wavelength.