Chapter 24, Problem 33P

To determine

The wavelengths of given angles.

Answer to Problem 33P

Wavelength for angle $31.2°$ is $534nm$ .

Wavelength for angle $36.4°$ is $612nm$ .

Wavelength for angle $47.5°$ is $760nm$ .

Explanation of Solution

Given info:

Angles, ${\theta }_1=31.2°,{\theta }_2=36.4°,{\theta }_3=47.5°$

Number of falling light per unit length, $N=9700lines/cm$

Formula used:

  $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer

  $\lambda$ is wavelength

  $d$ is distance between slits

Calculation:

In constructive interference, path difference is an integer multiple of wavelength.

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

Calculation for angle ${\theta }_1=31.2°$ by using $d\sin {\theta }_1=m{\lambda }_1$ where $m=1$

Substitute the given values in $d\sin {\theta }_1=m{\lambda }_1$

  $\left[\frac{1}{\left(9700lines/cm\right)}\right]\left({10}^{-2}m/cm\right)\sin 31.2°=\left(1\right)\left({\lambda }_1\right)$

  ${\lambda }_1=5.34\times {10}^{-7}m=534nm$

Calculation for angle ${\theta }_2=36.4°$ by using $d\sin {\theta }_2=m{\lambda }_2$ where $m=1$

Substitute the given values in $d\sin {\theta }_2=m{\lambda }_2$

  $\left[\frac{1}{\left(9700lines/cm\right)}\right]\left({10}^{-2}m/cm\right)\sin 36.4°=\left(1\right)\left({\lambda }_2\right)$

  ${\lambda }_2=6.12\times {10}^{-7}m=612nm$

Calculation for angle ${\theta }_3=47.5°$ by using $d\sin {\theta }_3=m{\lambda }_3$ where $m=1$

Substitute the given values in $d\sin {\theta }_3=m{\lambda }_3$

  $\left[\frac{1}{\left(9700lines/cm\right)}\right]\left({10}^{-2}m/cm\right)\sin 47.5°=\left(1\right)\left({\lambda }_3\right)$

  ${\lambda }_3=7.60\times {10}^{-7}m=760nm$

Conclusion:

The required wavelengths are $534nm,612nm,760nm$ .