Chapter 33, Problem 69P

(a)

To determine

The distance of first and second order intensity from the central intensity maximum.

Explanation of Solution

Given:

The wavelength of light is $589\text{nm}$ .

The diffraction grating has $2.0\text{cm}$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50\text{m}$ .

The focal length of lens is $1.50\text{m}$ .

Formula used:

Write the expression for diffraction pattern.

  $d\sin \theta =m\lambda$ …… (1)

Here, $d$ is the distance between two slits, $\theta$ is the angle at which diffraction occurs, $m$ is the order of diffraction and $\lambda$ is the wavelength of light.

Write the expression for position of intensity maxima.

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

Here, $y_m$ is the intensity maximum and $L$ is the distance of screen from grating.

Write the expression for slit separation.

  $d=\frac{1}{N}$ …… (2)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

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

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

Rearrange above expression for $y_m$ .

  $\begin{array}{c}\frac{d^2{y}_m^2}{\left(L^2+y_m^2\right)}=m^2{\lambda }^2\\ d^2{y}_m^2=m^2{\lambda }^2\left(L^2+y_m^2\right)\\ y_m^2\left(d^2-m^2{\lambda }^2\right)=m^2{\lambda }^2{L}^2\end{array}$

  $y_m=\frac{m\lambda L}{\sqrt{d^2-m^2{\lambda }^2}}$ …… (3)

Substitute $4000$ for $N$ in equation (2).

  $\begin{array}{c}d=\frac{1\text{cm}}{4000}\\ =0.00025\text{cm}\end{array}$

Substitute $589\text{nm}$ for $\lambda$ , $1.50\text{m}$ for $L$ and $0.00025\text{cm}$ for $d$ in equation (3).

  $\begin{array}{c}{y}_m=\frac{m\left(589\text{nm}\right)\left(1.50\text{m}\right)}{\sqrt{{\left(0.00025\text{cm}\right)}^2-m^2{\left(589\text{nm}\right)}^2}}\\ =\frac{m\left(589\text{nm}\right)\left(\frac{1\text{cm}}{{10}^7\text{nm}}\right)\left(1.50\text{m}\right)\left(\frac{1\text{cm}}{{10}^{-2}\text{m}}\right)}{\sqrt{{\left(0.00025\text{cm}\right)}^2-m^2{\left(589\text{nm}\right)}^2{\left(\frac{1\text{cm}}{{10}^7\text{nm}}\right)}^2}}\\ =\frac{m\left(883.5\times {10}^{-5}{\text{cm}}^2\right)}{\sqrt{\left(6.25\times {10}^{-8}{\text{cm}}^2\right)-m^2\left(0.347\times {10}^{-8}{\text{cm}}^2\right)}}\end{array}$

Substitute $1$ for $m$ (for first maximum) in above equation.

  $\begin{array}{c}{y}_m=\frac{\left(1\right)\left(883.5\times {10}^{-9}{\text{cm}}^2\right)}{\sqrt{\left(6.25\times {10}^{-8}{\text{ cm}}^2\right)-{\left(1\right)}^2\left(0.347\times {10}^{-8}{\text{cm}}^2\right)}}\\ =\frac{883.5\times {10}^{-5}{\text{cm}}^2}{2.43\times {10}^{-4}\text{cm}}\\ =36.3\text{cm}\end{array}$

Substitute $2$ for $m$ (for second maximum) in above equation.

  $\begin{array}{c}{y}_m=\frac{\left(2\right)\left(883.5\times {10}^{-9}{\text{cm}}^2\right)}{\sqrt{\left(6.25\times {10}^{-8}{\text{cm}}^2\right)-{\left(2\right)}^2\left(0.347\times {10}^{-8}{\text{cm}}^2\right)}}\\ =\frac{883.5\times {10}^{-5}{\text{cm}}^2}{2.20\times {10}^{-4}\text{cm}}\\ =40.1\text{cm}\end{array}$

Conclusion:

Thus, the distances of first and second maxima are $36.3\text{cm}$ and $40.1\text{cm}$ .

(b)

To determine

The width of central maxima.

Explanation of Solution

Given:

The wavelength of light is $589\text{nm}$ .

The diffraction grating has $2.0\text{cm}$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50\text{m}$ .

The focal length of lens is $1.50\text{m}$ .

Formula used:

Write the expression for minima of diffraction grating.

  $\begin{array}{c}{\theta }_{\min }=\frac{\lambda }{Nd}\\ =\frac{\Delta y}{2L}\end{array}$

Here, ${\theta }_{\min }$ is the minimum angle for first minima, $\lambda$ is the wavelength of light, $N$ is the number of lines in grating, $d$ is the slit separation, $\Delta y$ is the width of central maxima and $L$ is the distance between screen and grating.

Rearrange the above expression for $\Delta y$ .

  $\Delta y=\frac{2L\lambda }{Nd}$ …… (4)

Write the expression for slit separation.

  $d=\frac{1}{N}$ …… (5)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $4000$ for $N$ in equation (5).

  $\begin{array}{c}d=\frac{1\text{cm}}{4000}\\ =0.00025\text{cm}\end{array}$

Substitute $589\text{nm}$ for $\lambda$ , $1.50\text{m}$ for $L$ , $8000$ for $N$ and $0.00025\text{cm}$ for $d$ in equation (4).

  $\begin{array}{c}\Delta y=\frac{2\left(1.50\text{m}\right)\left(589\text{nm}\right)}{\left(8000\right)\left(0.00025\text{cm}\right)}\\ =\frac{2\left(1.50\text{m}\right)\left(589\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{\left(8000\right)\left(0.00025\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)}\\ =\frac{1767\times {10}^{-9}{\text{m}}^2}{2\times {10}^{-2}\text{m}}\left(\frac{1\text{μm}}{{10}^{-6}\text{m}}\right)\\ =88.4\text{μm}\end{array}$

Conclusion:

Thus, the width of central maxima is $88.4\text{μm}$ .

(c)

To determine

The resolution in first order.

Explanation of Solution

Given:

The wavelength of light is $589\text{nm}$ .

The diffraction grating has $2.0\text{cm}$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50\text{m}$ .

The focal length of lens is $1.50\text{m}$ .

Formula used:

Write the expression for resolution power.

  $R=mN$ ……. (6)

Here, $R$ is the resolving power of grating, $m$ is the order of spectrum and $N$ is the total number of lines.

Calculation:

Substitute $1$ for $m$ and $8000$ for $N$ in equation (6).

  $\begin{array}{c}R=\left(1\right)\left(8000\right)\\ =8000\end{array}$

Conclusion:

Thus, the resolution power of grating is $8000$ .