Chapter 19, Problem 61A

(a)

To determine

The number of grooves present in a centimeter along the radius of the record.

Answer to Problem 61A

The number of the grooves present along the radius of the record is $83\text{ridges/cm}$ .

Explanation of Solution

Given:

The distance of the screen from the slit is $L=4.0\text{m}$ .

The separation of lines in diffraction pattern is $x=21\text{mm}$ .

The wavelength of the laser is $\lambda =632.8\text{nm}$ .

The diffraction grating of the record is $33\frac{1}{3}$ rpm or $\left(\frac{100}{3}\right)$ rpm.

The given figure is shown below.

  

Formula used:

The expression for the distance between two ridges is,

  $\begin{array}{l}\lambda =d\sin \theta \\ d=\frac{\lambda }{\sin \theta }\\ d=\frac{\lambda }{\sin \left[ta{n}^{-1}\left(\frac{x}{L}\right)\right]}\end{array}$

Calculation:

The distance $\left(d\right)$ between the two ridges is,

  $\begin{array}{l}d=\frac{\lambda }{\sin \left[ta{n}^{-1}\left(\frac{x}{L}\right)\right]}\\ d=\frac{632.8\text{nm}\times \left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{\sin \left[ta{n}^{-1}\left(\frac{21.0\text{mm}\times \frac{1\text{m}}{1000\text{mm}}}{4.0\text{m}}\right)\right]}\\ d=\frac{632.8\times {10}^{-9}\text{m}}{\sin \left[ta{n}^{-1}\left(\frac{0.021\text{m}}{4.0\text{m}}\right)\right]}\\ d=1.20\times {10}^{-4}\text{m}\times \left(\frac{100\text{cm}}{1\text{m}}\right)\\ d=1.20\times {10}^{-2}\text{cm}\end{array}$

The number of ridges $\left(n\right)$ are

  $\begin{array}{l}n=\frac{1}{d}\\ n=\left(\frac{1}{1.20\times {10}^{-2}\text{cm}}\right)\\ n=83\text{ridges/cm}\end{array}$

Conclusion:

Thus, the number of the grooves present along the radius of the record is $83\text{ridges/cm}$ .

(b)

To determine

The number of grooves present in a centimeter.

Answer to Problem 61A

The number of grooves is present in a centimeter, when the song takes $16\text{mm}$ on the record and last for $4.01\text{min}$ is $83\text{ridges}$ .

Explanation of Solution

Given:

The diffraction grating of the record is $dg=33\frac{1}{3}$ rpm or $\left(\frac{100}{3}\right)$ rpm.

A song takes $16\text{mm}$ on the record and last for $4.01\text{min}$

The given figure is shown below.

  

Formula used:

The number of ridges are:

  $m=dg\times t$

Here, dg is the diffraction grating.

Calculation:

The number $\left(m\right)$ of ridges that will last for $t=4.01\text{min}$ are,

  $\begin{array}{l}m=\left(dg\right)\times t\\ m=\left(\frac{100}{3}\text{ridges/min}\right)\times 4.01\text{min}\\ m=133.67\text{ridges}\end{array}$

The $133.67$

  $\text{ridges}$ take $\left(16\text{mm}\times \frac{1\text{cm}}{10\text{mm}}\right)=1.6\text{cm}$ of the record

The number of ridges present in a centimeter of the record are,

  $\begin{array}{l}\text{Number}\text{of}\text{ridges}=\left[\frac{133.67\text{ridges}}{16\text{mm}\times \left(\frac{1\text{cm}}{10\text{mm}}\right)}\right]\\ \text{Number}\text{of}\text{ridges}=\left(\frac{133.67\text{ridges}}{1.6\text{cm}}\right)\\ \text{Number}\text{of}\text{ridges}=83\text{ridges}\end{array}$

Conclusion:

Thus, the number of grooves is present in a centimeter, when the song takes $16\text{mm}$ on the record and last for $4.01\text{min}$ is $83\text{ridges}$ .