Chapter 37, Problem 45AP

(a)

To determine

To show: The dispersion is given by $\frac{d\lambda }{dy}=\frac{L^{2}d}{m{\left(L^2+y^2\right)}^{\frac{3}{2}}}$.

Answer to Problem 45AP

The dispersion is, $\frac{d\lambda }{dy}=\frac{L^{2}d}{m{\left(L^2+y^2\right)}^{\frac{3}{2}}}$.

Explanation of Solution

Formula to calculate the angles of bright beams diffracted from the grafting is,

$d\sin \theta =m\lambda$ (1)

Here,

$d$ is the spacing between adjacent slits.

$m$ is order number of intensity maximum.

$\lambda$ is wavelength of light.

$\theta$ is the angle by which ray is diffracted.

Write the expression for sine of angle $\theta$.

$\sin \theta =\frac{y}{\sqrt{L^2+y^2}}$

Substitute $\frac{y}{\sqrt{L^2+y^2}}$ for $\sin \theta$ in equation (1)

$d\frac{y}{\sqrt{L^2+y^2}}=m\lambda$

Differentiate the above equation with respect to $y$.

$\frac{d}{dy}\left(d\frac{y}{\sqrt{L^2+y^2}}\right)=\frac{d}{dy}\left(m\lambda \right)$

Apply product rule of differentiation to differentiate above equation.

$\begin{array}{c}d{\left(L^2+y^2\right)}^{-\frac{1}{2}}+\left(d\right)y\left(-\frac{1}{2}\right){\left(L^2+y^2\right)}^{-\frac{3}{2}}\left(0+2y\right)=m\frac{d\lambda }{dy}\\ \frac{d}{{\left(L^2+y^2\right)}^{\frac{1}{2}}}-\frac{\left(d\right)y^2}{{\left(L^2+y^2\right)}^{\frac{3}{2}}}=m\frac{d\lambda }{dy}\\ \frac{\left(d\right)\left(L^2+y^2\right)-\left(d\right)y^2}{{\left(L^2+y^2\right)}^{\frac{3}{2}}}=m\frac{d\lambda }{dy}\\ \frac{d\lambda }{dy}=\frac{L^{2}d}{m{\left(L^2+y^2\right)}^{\frac{3}{2}}}\end{array}$

Conclusion:

Therefore, the dispersion is $\frac{d\lambda }{dy}=\frac{L^{2}d}{m{\left(L^2+y^2\right)}^{\frac{3}{2}}}$.

(b)

To determine

The dispersion in first order.

Answer to Problem 45AP

The dispersion in first order is $3.77\text{nm}/\text{cm}$.

Explanation of Solution

Given info: The mean wavelength of light is $550\text{nm}$, grating is $8000\text{ruling}/\text{cm}$, and screen is placed at a distance of $2.40\text{m}$.

Formula to calculate the angles of bright beams diffracted from the grafting is,

$d\sin \theta =m\lambda$ (2)

Here,

$d$ is the spacing between adjacent slits.

$m$ is order number of intensity maximum.

$\lambda$ is wavelength of light.

$\theta$ is the angle by which ray is diffracted.

The spacing between adjacent slit is inverse of number of rulings per centimeter is,

$\begin{array}{c}d=\frac{1}{8000}\text{cm}\\ \text{=1}\text{.25}\times {\text{10}}^{-4}\text{cm}\end{array}$

Substitute $\text{1}\text{.25}\times {\text{10}}^{-4}\text{cm}$ for $d$, 1 for $m$ and $550\text{nm}$ for $\lambda$ in equation (2).

$\begin{array}{c}\text{1}\text{.25}\times {\text{10}}^{-4}\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\left(\sin \theta \right)=1\cdot 550\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\\ \text{1}\text{.25}\times {\text{10}}^{-6}\text{m}\cdot \text{sin}\theta =550\times {10}^{-9}\text{m}\\ \theta ={\sin }^{-1}\left(\frac{550\times {10}^{-9}\text{m}}{\text{1}\text{.25}\times {\text{10}}^{-6}\text{m}}\right)\\ =26.10°\end{array}$

For the value of $y$,

$\begin{array}{c}\tan \theta =\frac{y}{L}\\ y=L\tan \theta \end{array}$

Substitute $2.40\text{m}$ for $L$, and $26.1°$ for $\theta$ in above equation.

$\begin{array}{c}y=2.40\tan \left(26.1°\right)\\ =2.40\text{m}\left(0.489\right)\\ =1.18\text{m}\end{array}$

Formula to calculate the dispersion is,

$\frac{d\lambda }{dy}=\frac{L^{2}d}{m{\left(L^2+y^2\right)}^{\frac{3}{2}}}$

Here,

$L$ is the distance between slits and screen.

$m$ is order number of intensity maximum.

$y$ is position relative to the center of a diffraction pattern.

Substitute $2.40\text{m}$ for $L$, 1 for $m$, $1.18\text{m}$ for $y$ and $\text{1}\text{.25}\times {\text{10}}^{-6}\text{m}$ for $d$ to calculate $\frac{d\lambda }{dy}$.

$\begin{array}{c}\frac{d\lambda }{dy}=\frac{{\left(2.4\text{m}\right)}^2\left(1.25\times {10}^{-6}\text{m}\right)}{1{\left({\left(2.4\text{m}\right)}^2+{\left(1.18\text{m}\right)}^2\right)}^{\frac{3}{2}}}\\ =\frac{7.2\times {10}^{-6}\text{m}}{{\left(7.1524\right)}^{\frac{3}{2}}\text{m}}\\ =3.77\frac{{10}^{-7}\text{m}}{\text{1m}\times \frac{{\text{10}}^2\text{cm}}{1\text{m}}}\\ =3.77\frac{{10}^{-9}\text{m}}{\text{cm}}\end{array}$

Further solve the above equation.

$\begin{array}{c}\frac{d\lambda }{dy}=\frac{3.77\times {10}^{-9}\text{m}\times \left(\frac{{\text{10}}^9\text{nm}}{1\text{m}}\right)}{\text{cm}}\\ =3.77\text{nm}/\text{cm}\end{array}$

Conclusion:

Therefore, the dispersion is $3.77\text{nm}/\text{cm}$.