Question

A He–Ne laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an angle equal to the angle of incidence. However, fringes are also observed. If the wall is 1.50 m from the CD, and the first fringe is 0.600 m from the central maximum, what is the spacing of grooves on the CD?

Short answer

The spacing of grooves on the CD is $6.82\times {10}^{‐7}\mathrm{m}$.

Step-by-step solution

Concept Introduction

A laser beam is a stream of focused, coherent waves in a single wavelength.

Given data

The distance between the wall and the CD is x = 1.50 m.

The distance of the first fringe from the central maximum is y = 0.600 m

The wavelength of the He-Ne laser is $\mathrm{\lambda }=633\hspace{0.33em}\mathrm{nm}\left(\frac{{10}^{‐9}\mathrm{m}}{1\mathrm{nm}}\right)=6.33\times {10}^{‐7}\hspace{0.33em}\mathrm{m}$

Determine the spacing of grooves on the CD

The equation for the constructive interference can be given as

$\mathrm{dsin\theta }=\mathrm{m\lambda }$ …………………..(1)

Using the distance of x and y we can calculate the angle from the equation,

$$\begin{gathered} \mathrm{\theta }={\tan }^{‐1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right) \\ ={\tan }^{‐1}\left(\frac{1.50\mathrm{m}}{0.600\mathrm{m}}\right) \\ =68.2° \end{gathered}$$

Substitute these values in the equation (1), and we get,

$$\begin{gathered} \mathrm{d}=\frac{\mathrm{m\lambda }}{\mathrm{sin\theta }} \\ =\frac{1\times 6.33\times {10}^{‐7}\mathrm{m}}{\sin \left(68.2°\right)} \\ =6.82\times {10}^{‐7}\mathrm{m} \end{gathered}$$

Therefore, the spacing of grooves on the CD is $=6.82\times {10}^{‐7}\mathrm{m}$.