91Ó°ÊÓ

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,$\mathrm{λ}$.

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.

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,

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

.......(1)

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

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

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

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

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

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

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

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

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