Question

In Fig. 35-31, a light wave along ray ${\mathit{r}}_{\mathbf{1}}$reflects once from a mirror and a light wave along ray ${\mathit{r}}_{\mathbf{2}}$reflects twice from that same mirror and once from a tiny mirror at distance $\mathit{L}$from the bigger mirror. (Neglect the slight tilt of the rays.) The waves have wavelength $\mathit{\lambda }$and are initially exactly out of phase. What are the (a) smallest (b) second smallest, and (c) third smallest values of $\frac{\mathbf{L}}{\mathbf{\lambda }}$that result in the final waves being exactly in phase?

Short answer

1. The smallest value of $\frac{L}{\lambda }$is$\frac{1}{4}$
2. The second smallest value of $\frac{L}{\lambda }$is $\frac{3}{4}$.
3. The third smallest value of$\frac{L}{\lambda }$ is$\frac{5}{4}$ .

Step-by-step solution

Given information

The ray reflected once from the bigger mirror is,$r_1$ .

The ray reflected twice from the bigger mirror and once from a tiny mirror is, $r_2$.

The distance between the tiny mirror and bigger mirror is, $L$.

The wavelength of both the rays is,$\lambda$.

Path difference of waves

The two light waves moving with some initial phase difference between them can result inthe differentvalue of phase difference. It means that, the waves travel through paths having different lengths beforecomingback together.

The value of the path length difference between two light waves changes with the change in the value of the wavelength.

 a) The smallest value

For the wave$r_2$, the value of the path distance travelled by wave $r_2$is given by,

$d_1=4L$

For the wave $r_1$,the value of the path distance travelled by wave $r_1$is given by,

$d_2=2L$

When they are out of phase then the equation for the value ofpath difference is given by,

$\begin{array}{rcl}{d}_1-d_2& =& \left(n+\frac{1}{2}\right)\lambda \\ 4L-2L& =& \left(n+\frac{1}{2}\right)\lambda \\ 2L& =& \left(n+\frac{1}{2}\right)\lambda \\ \frac{L}{\lambda }& =& \frac{1}{2}\left(n+\frac{1}{2}\right)\end{array}$

.......(1)

Putting $n=0$in equation (1), the smallest value of $\frac{L}{\lambda }$is given by,

$$\begin{gathered} \frac{L}{\lambda }=\frac{1}{2}\left(0+\frac{1}{2}\right) \\ \frac{L}{\lambda }=\frac{1}{2}\left(\frac{1}{2}\right) \\ \frac{L}{\lambda }=\frac{1}{4} \end{gathered}$$

Hence, the smallest value of $\frac{L}{\lambda }$is $\frac{1}{4}$.

(b) The second smallest value

Putting $n=0$in equation (1), the second smallest value of $\frac{L}{\lambda }$is given by,

$$\begin{gathered} \frac{L}{\lambda }=\frac{1}{2}\left(1+\frac{1}{2}\right) \\ \frac{L}{\lambda }=\frac{1}{2}\left(\frac{3}{2}\right) \\ \frac{L}{\lambda }=\frac{3}{4} \end{gathered}$$

Hence, the second smallest value of $\frac{L}{\lambda }$is$\frac{3}{4}$ .

(c) The third smallest value

Putting $n=2$in equation (1), the third smallest value of $\frac{L}{\lambda }$is given by,

$$\begin{gathered} \frac{L}{\lambda }=\frac{1}{2}\left(2+\frac{1}{2}\right) \\ \frac{L}{\lambda }=\frac{1}{2}\left(\frac{5}{2}\right) \\ \frac{L}{\lambda }=\frac{5}{4} \end{gathered}$$

Hence, the third smallest value of $\frac{L}{\lambda }$is$\frac{5}{4}$ .