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 620 nm and are initially in phase. (a) What is the smallest value of $\mathit{L}$that puts the final light waves exactly out of phase? (b) With the tiny mirror initially at that value of $\mathit{L}$, how far must it be moved away from the bigger mirror to again put the final waves out of phase?

Short answer

1. The smallest value of that puts the final light waves exactly out of phase is $155\text{nm}$.
2. The tiny mirror should be moved $310\text{nm}$away from the bigger mirror to again put the final waves out of phase.

Step-by-step solution

Given information

The wavelength of both light rays is,$\lambda =620\text{nm}$

Destructive interference

When two waves of equal frequency and opposite phase meet each other then they cancel each other out and results in the ‘destructive interference’.

For destructive interference, the value of the path difference between two waves is given by,

$\Delta x=\left(2n+1\right)\left(\frac{\lambda }{2}\right)$

Here,$\lambda$ is the wavelength of both waves and $n=1,2,3,...$is the number of fringes.

(a) Smallest value of distance between tiny and bigger mirror

According to the question, the wave$W_2$reflects twice from that same mirror and once from a tiny mirror located at a distance $L$. Has no substantive effect on the calculations,

The value of the phase difference after two reflections will be,

$2\left(\frac{\lambda }{2}\right)=\lambda$

The distance travelled by wave $W_2$is $2L$greater than the distance travelled by wave $W_1$.

Then the smallest value of $L$that puts the wave role="math" localid="1663049389226" $W_2$a half-wavelength "behind" wave $W_1$is given by,

$$\begin{gathered} 2L=\frac{\lambda }{2} \\ L=\frac{\lambda }{4} \\ L=\frac{620\text{nm}}{4} \\ L=155\text{nm} \end{gathered}$$

Hence, the smallest value of$L$ that puts the final light waves exactly out of phase is$155\text{nm}$ .

(b) Distance moved by tiny mirror

The fact that wave $W_2$reflects two additional times has no substantive effect on the calculations. Since two reflections amount to a $2\left(\frac{\lambda }{2}\right)=\lambda$phase difference, which is effectively not a phase difference at all. The substantive difference between $W_2$and $W_1$is the extra distance 2L travelled by $W_2$. Destructive interference will again appear if$W_2$is$\frac{3\lambda }{2}$"behind" the other wave.

In this case, $2L\text{'}=\frac{3\lambda }{2}$and the difference is,

$$\begin{gathered} L\text{'}-L=\frac{3\lambda }{4}-\frac{\lambda }{4} \\ L\text{'}-L=\frac{\lambda }{2} \\ L\text{'}-L=\frac{620\text{nm}}{2} \\ L\text{'}-L=310\text{nm} \end{gathered}$$

Hence, the tiny mirror should be moved $310\text{nm}$away from the bigger mirror to again put the final waves out of phase.