Chapter 24, Problem 32P

To determine

The angular spread in second order spectrum between given wavelengths.

Answer to Problem 32P

  $24°$

Explanation of Solution

Given info:

Wavelength of red, ${\lambda }_R=7\times {10}^{-7}m$

Wavelength of blue, ${\lambda }_B=4.5\times {10}^{-7}m$

Number of grating per unit length $N=6.0\times {10}^{5}lines/cm$

Formula used:

  $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer.

  $\lambda$ is wavelength.

  $d$ is distance between slits.

Calculation:

The path difference is an integer multiple of the wavelength in constructive interference.

Rearrange the given formula: $\sin \theta =\frac{m\lambda }{d}$ . So, $d\sin \theta =m\lambda$

Calculation for Wavelength of red by using $d\sin \theta =m\lambda$ with $m=2$

Substitute the given values in $d\sin \theta =m\lambda$

  $\frac{1}{6.0\times {10}^{5}lines/m}\sin {\theta }_1=\left(2\right)\left(7.0\times {10}^{-7}m\right)$

  $\sin {\theta }_1=0.84$

  ${\theta }_1=57.1°$

Calculation for Wavelength of blue by using $d\sin \theta =m\lambda$ with $m=2$

Substitute the given values in $d\sin \theta =m\lambda$

  $\frac{1}{6.0\times {10}^{5}lines/m}\sin {\theta }_2=\left(2\right)\left(4.5\times {10}^{-7}m\right)$

  $\sin {\theta }_2=0.54$

  ${\theta }_2=32.7°$

Therefore, the angular spread in the second order spectrum between both wavelengths:

  $\Delta \theta =57.1°-32.7°=24°$

Conclusion:

The required angle is $24°$ .