91Ó°ÊÓ

Wavelength is defined as the distance between identical points in the adjacent cycles of waveform signal propagated in space or along a wire.

Assume the wedge-shaped film is in air, so the wave reflected from one surface undergoes a phase charge of $\mathrm{Ï€}$ rad while the wave reflected from the other surface does not. At a place where the film thickness is $L$ he condition of fully destructive interference is

$2L=\frac{\mathrm{λ}}{n}\left(m=0,1,2.....\right)$

To find the thickness difference $\mathrm{Δ}L$ between left and right end, twice to go throw above equation, once for the thickness $L_L$at the left end and once for the thickness $L_R$ at the right end.

$L_L=m_L\frac{\mathrm{λ}}{2n}$

And

$L_R=\left(m_L+9\right)\frac{\mathrm{λ}}{2n}$

Where $m_L$ be the value at the left end for which dark fringe is observed. Then the value of the right end must be$m_L+9$ because, the right end s located at the ninth dark fringe from the left end.

Wavelength of the incident light $\mathrm{λ}=630\text{nm}$

$$\begin{gathered} \left(630\text{mm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right) \\ 630×{10}^{-9}\text{m} \end{gathered}$$

Refractive index of the film $n=150$

Subtracting the film thickness o the left $L_L$and right $L_R$ we get

$$\begin{gathered} \mathrm{Δ}L=L_R-L_L \\ =\left(m_L+9\right)\frac{\mathrm{λ}}{2\mathrm{π}} \\ \mathrm{Δ}L=\frac{9\mathrm{π}}{2n}..........\left(1\right) \end{gathered}$$

From the given data, subtracting the value of wavelength $\mathrm{λ}=630×{10}^{-9}\text{m}$ and refractive index $\text{n = 1.5}$ in the above equation (1), we get

$$\begin{gathered} \mathrm{Δ}L=\frac{9\mathrm{π}}{2n} \\ =\frac{9\left(630×{10}^{-9}\text{m}\right)}{2\left(1.5\right)} \\ =1.89×{10}^{-6}\text{m} \\ \text{= 1.89}\mathrm{μ}\text{m} \end{gathered}$$

Therefore, from left-to right change in thin film thickness is$1.89\mathrm{μ}\text{m}$