Chapter 23, Problem 32E

To determine

Position of the first line.

Answer to Problem 32E

The position of the first line from the centre of the screen is $0.24\text{m}$.

Explanation of Solution

Write the expression for diffraction grating.

  $d\sin \theta =m\lambda$        (I)

Here, $\lambda$ is the wavelength of light, $m$ is the order of interference, $d$ is the distance between the slits and $\theta$ is the angle of the beam.

When $\theta$ is small $\sin \theta \approx \theta$ the above equation can be written as

  $d\cdot \theta =m\lambda$        (II)

Write the expression for angle of the beam.

  $\theta =\frac{y}{\sqrt{y^2+D^2}}$        (III)

Here, $y$ is the position of the fringe and $D$ is the distance between the source and screen.

Substitute equation (III) in equation (II)

  $\begin{array}{l}\frac{y}{\sqrt{y^2+D^2}}=\frac{m\lambda }{d}\\ \frac{y^2}{y^2+D^{{}^2}}={\left(\frac{m\lambda }{d}\right)}^2\\ y^2{d}^2={\left(m\lambda \right)}^2\left(y^2+D^{{}^2}\right)\\ y^2{d}^2=m^2{\lambda }^2{y}^2+D^2{m}^2{\lambda }^2\end{array}$

  $\begin{array}{l}\left(y^2{d}^2-m^2{\lambda }^2{y}^2\right)=D^2{m}^2{\lambda }^2\\ y^2\left(d^2-m^2{\lambda }^2\right)=D^2{m}^2{\lambda }^2\\ y^2=\frac{D^2{m}^2{\lambda }^2}{\left(d^2-m^2{\lambda }^2\right)}\end{array}$

  $y={\left(\frac{D^2{m}^2{\lambda }^2}{\left(d^2-m^2{\lambda }^2\right)}\right)}^{\frac{1}{2}}$        (IV)

Write the expression for space between the two adjacent lines.

  $d=\frac{1}{N}$        (V)

Here, $N$ is the number of lines per centimetre in the grating.

Conclusion:

Substitute $8000\text{cm}$ for $N$ in equation (V) to find $d$

  $\begin{array}{c}d=\frac{1}{8000\text{cm}}\\ =1.25\times {10}^{-4}\text{cm}\end{array}$

The distance between the two adjacent lines is $1.25\times {10}^{-4}\text{m}$

Substitute $1.25\times {10}^{-4}\text{cm}$ for $d$, $0.5\text{m}$ for $D$, $550\text{nm}$ for $y$, $1$ for $m$ in equation (IV) to find $y$

  $\begin{array}{c}y={\left(\frac{{\left(0.5\text{m}\right)}^2{\left(1\right)}^2{\left(550\times {10}^{-9}\text{m}\right)}^2}{\left({\left(1.25\times {10}^{-4}\text{cm}\left(\frac{\text{0}\text{.01}\text{m}}{\text{1}\text{cm}}\right)\right)}^2-{\left(1\right)}^2{\left(550\times {10}^{-9}\text{m}\right)}^2\right)}\right)}^{\frac{1}{2}}\\ ={\left(\frac{\left(0.25{\text{m}}^2\right)\left(3.025\times {10}^{-13}{\text{m}}^2\right)}{{\left(1.25\times {10}^{-6}\text{m}\right)}^2-\left(3.025\times {10}^{-13}{\text{m}}^2\right)}\right)}^{\frac{1}{2}}\\ ={\left(\frac{\left(\text{7}{\text{.5625×10}}^{\text{-14}}{\text{m}}^4\right)}{\left(\text{1}{\text{.5625×10}}^{\text{-12}}{\text{m}}^{\text{2}}\right)\text{-}\left(\text{3}{\text{.025×10}}^{\text{-13}}{\text{m}}^{\text{2}}\right)}\right)}^{\frac{1}{2}}\\ ={\left(\frac{\left(7.5625\times {10}^{-14}{\text{m}}^4\right)}{\left(1.2601\times {10}^{-12}{\text{m}}^2\right)}\right)}^{\frac{1}{2}}\\ ={\left(0.060{\text{m}}^{\text{2}}\right)}^{\frac{1}{2}}\\ =0.24\text{m}\end{array}$

Therefore, the position of the first line from the centre of the screen is $0.24\text{m}$.