Question

A thin plastic sheet of refractive index $1.6$ is used to cover one of the slits of a double slit arrangement. The central point on the screen is now occupied by what would have been the ${{7}^{th}}$ bright fringe before the plastic was used. If the wavelength of the light is $600nm$, what is the thickness $\left( \text{in }\mu m \right)$ of the plastic.
$A)\text{ }7$
$B)\text{ 4}$
$C)\text{ 8}$
$D)\text{ 6}$

Answer 1

Hint : This problem can be solved by calculating the change in the path difference that is caused due to the introduction of the thin plastic sheet in terms of its thickness. This change in path difference can then be equated to the numerical value for the change in path difference that is required for the central bright fringe to be in the place of the seventh bright fringe if the sheet was not there.
Formula used:
$\Delta x=\left( \mu -1 \right)t$

$\Delta x=n\lambda$

Complete step by step answer:
To solve this question, we will find the extra path difference caused due to the introduction of the plastic sheet and equate it with the extra path difference that is required to shift the central bright fringe to the position of the seventh bright fringe.
Hence, let us proceed to do that.
Let the thickness of the plastic sheet be $t$.
The refractive index of the sheet is $\mu =1.6$.
The wavelength of the light used is $\lambda =600nm$.
Let the extra path difference that is caused due to the introduction of the sheet in front of one of the slits be $\Delta {{x}_{sheet}}$.
Now, the extra path difference $\Delta x$ that is caused due to the introduction of a sheet of refractive index $\mu$ in one of the slits of the Young’s Double Slit Experiment is given by,
$\Delta x=\left( \mu -1 \right)t$ --(1)
where $t$ is the thickness of the sheet.
Hence, using (1), we get,
$\Delta {{x}_{sheet}}=\left( \mu -1 \right)t=\left( 1.6-1 \right)t=0.6t$ --(2)
Now, let the path difference for the formation of the seventh bright fringe be $\Delta {{x}_{7}}$.
Let the path difference for the formation of the central bright fringe be $\Delta {{x}_{0}}$.
The path difference $\Delta x$ in YDSE for the ${{n}^{th}}$ bright fringe is given by
$\Delta x=n\lambda$ --(3)
where $\lambda$ is the wavelength of the light used.
Using (3), we get,
$\Delta {{x}_{7}}=7\lambda$ $\left( \because \text{For }{{\text{7}}^{th}}\text{ bright fringe, }n=7 \right)$
$\Delta {{x}_{0}}=0\lambda =0$ $\left( \because \text{For central bright fringe, }n=0\text{ } \right)$
Therefore, the extra path difference $\Delta {{x}_{required}}$ required to shift the central bright fringe to the position of the seventh bright fringe will be
$\Delta {{x}_{required}}=\Delta {{x}_{7}}-\Delta {{x}_{0}}$
$\therefore \Delta {{x}_{required}}=7\lambda -0=7\lambda$ --(4)
Now, according to the question, the introduction of the thin sheet in front of the slit causes this extra path difference. Hence,
$\Delta {{x}_{required}}=\Delta {{x}_{sheet}}$ --(5)
Therefore, using (2) and (4) in (5), we get,
$7\lambda =0.6t$
$\therefore t=\dfrac{7\lambda }{0.6}=\dfrac{7\times 600\times {{10}^{-9}}}{0.6}$ $\left( \because 1nm={{10}^{-9}}m \right)$
$\therefore t=7000\times {{10}^{-9}}=7\times {{10}^{-6}}=7\mu m$ $\left( \because 1\mu m={{10}^{-6}}m \right)$
Therefore, the thickness of the sheet is $7\mu m$.
Therefore, the correct option is $A)\text{ }7$.

Note : These types of complex problems wherein either the apparatus is put in a medium of different refractive index or a sheet is put in either in front of either one or both the slits. However, these can be solved very easily just by calculating the path difference that is caused by the changes in the experiment. Calculating these path differences are usually very simple and have specific formulas for this purpose. By calculating the path difference and comparing with some condition given, such problems can be easily solved.