Chapter 24, Problem 36P

To determine

The width of the first order spectrum on a screen $\text{2}\text{.30}\text{m}$ away

Answer to Problem 36P

  $1.05m$ .

Explanation of Solution

Given info:

Number of grating per unit length, $N=8500lines/cm$

First Wavelength, ${\lambda }_1=410nm=410\times {10}^{-9}m$

Second Wavelength, ${\lambda }_2=750nm=750\times {10}^{-9}m$

Length, $L=2.30m$

Formula used:

  $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer.

  $\lambda$ is wavelength.

  $d$ is distance between the 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 $\sin {\theta }_{410}$ by using $d\sin {\theta }_{410}=m\lambda$ where, $m=1\left(firstorder\right)$

  $\left[\frac{1}{\left(8500lines/m\right)}\right]\left({10}^{-2}m/cm\right)\sin {\theta }_{410}=\left(1\right)\left(410\times {10}^{-9}m\right)$

  $\sin {\theta }_{410}=0.3485$

  ${\theta }_{410}=20.4°$

Calculation for $\sin {\theta }_{750}$ by using $d\sin {\theta }_{750}=m\lambda$ where, $m=1\left(firstorder\right)$

  $\left[\frac{1}{\left(8500lines/m\right)}\right]\left({10}^{-2}m/cm\right)\sin {\theta }_{750}=\left(1\right)\left(750\times {10}^{-9}m\right)$

  $\sin {\theta }_{750}=0.6375$

  ${\theta }_{750}=39.6°$

Now, calculate the distances from the central white line on the screen:

  $y_{410}=L\tan {\theta }_{410}=\left(2.30m\right)\tan 20.4°=0.855m$

  $y_{750}=L\tan {\theta }_{750}=\left(2.30m\right)\tan 39.6°=1.90m$

Therefore, the width of the spectrum:

  $y_{750}-y_{410}=1.90m-0.855m=1.05m$

Conclusion:

The required width is $1.05m$ .