A double-slit experiment is set up using a helium-neon laser $(\lambda =633\mathrm{nm})$. Then a very thin piece of glass $(n=1.50)$ is placed over one of the slits. Afterward, the central point on the screen is occupied by what had been the $m=10$ dark fringe. How thick is the glass?

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The passage of one ray over the other will be extended by inserting the glass. The number of wavelengths inside the glass can be stated as if the glass is thick.

$m_{\text{glass}}=\frac{d}{\lambda /n}=\frac{nd}{\lambda }$

here $n$is the glass's refractive index. We still must find that the amount of lengths in the air at height in order to calculate the change in path-length difference after inserting the glass. $\left(d\right)$"before adding the glass," then deduct that proportion from the diamond's excitation wavelength. The total number of wavelengths in the air equals the total number of wavelengths in the air.

$m_{\mathrm{air}}=\frac{d}{\lambda }$

After installing the glass, the difference in phase difference is now

$$\begin{gathered} m_{glass}-m_{\text{air}}=\frac{nd}{\lambda }-\frac{d}{\lambda } \\ =\frac{d}{\lambda }(n-1) \end{gathered}$$

Keep in mind that we count the fringes from zero, so the first fringe has $m=0$and the tenth fringe has $m=9$. When the 10th dark fringe moves to the situation of the first dark fringe, this equals $\mathrm{\Delta }m=9-0=9$, but when it moves to the location of the related topic, it equals $\mathrm{\Delta }m=9+0.5=9.5$, We still must find that the amount of lengths in the air at height in order to calculate the change in path-length difference after inserting the glass.

$9.5=\frac{d}{\lambda }(n-1)$

to determine $d$, adjust the equation

$$\begin{gathered} d=\frac{9.5\lambda }{(n-1)} \\ =\frac{9.5\times 0.633\mu \mathrm{m}}{1.5-1} \end{gathered}$$

$d=12\mu \mathrm{m}$