Optics 6 Question 49
51. The Young’s double slit experiment is done in a medium of refractive index 4/3. A light of $600 nm$ wavelength is falling on the slits having $0.45 mm$ separation. The lower slit $S _2$ is covered by a thin glass sheet of
thickness $10.4 \mu m$ and refractive index 1.5. The interference pattern is observed on a screen placed $1.5 m$ from the slits as shown in the figure.
(a) Find the location of central maximum (bright fringe with zero path difference) on the $y$-axis.
(b) Find the light intensity of point $O$ relative to the maximum fringe intensity.
(c) Now, if $600 nm$ light is replaced by white light of range 400 to $700 nm$, find the wavelengths of the light that form maxima exactly at point $O$.
(All wavelengths in the problem are for the given medium of refractive index $4 / 3$. Ignore dispersion)
$(1999,10 M)$
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Answer:
Correct Answer: 51. (a) $4.33 mm$ (b) $I=\frac{3 I _{\max }}{4}$ (c) $650 nm ; 433.33 nm$
Solution:
- Given, $\lambda=600 nm=6 \times 10^{-7} m$,
$$ \begin{aligned} d & =0.45 mm=0.45 \times 10^{-3} m, \\ D & =1.5 m . \end{aligned} $$
Thickness of glass sheet, $t=10.4 \mu m=10.4 \times 10^{-6} m$
Refractive index of medium, $\mu _m=4 / 3$
and refractive index of glass sheet, $\mu _g=1.5$
(a) Let central maximum is obtained at a distance $y$ below point $O$. Then $\Delta x _1=S _1 P-S _2 P=\frac{y d}{D}$
Path difference due to glass sheet
$$ \Delta x _2=\frac{\mu _g}{\mu _m}-1 t $$
Net path difference will be zero when
$$ \begin{array}{rlrl} \Delta x _1 & =\Delta x _2 \\ \text { or } \quad & \frac{y d}{D} & =\frac{\mu _g}{\mu _m}-1 t \\ \therefore \quad y & =\frac{\mu _g}{\mu _m}-1 t \frac{D}{d} \end{array} $$
Substituting the values, we have
$$ \begin{aligned} & y=\frac{1.5}{4 / 3}-1 \frac{10.4 \times 10^{-6}(1.5)}{0.45 \times 10^{-3}} \\ & y=4.33 \times 10^{-3} m \end{aligned} $$
or we can say $y=4.33 mm$.
(b) At $O, \Delta x _1=0$ and $\Delta x _2=\frac{\mu _g}{\mu _m}-1 t$
$\therefore$ Net path difference, $\Delta x=\Delta x _2$ Corresponding phase difference, $\Delta \varphi$
or $\quad$ simple $\varphi=\frac{2 \pi}{\lambda} . \Delta x$
Substituting the values, we have
$$ \begin{aligned} \varphi & =\frac{2 \pi}{6 \times 10^{-7}} \frac{1.5}{4 / 3}-1\left(10.4 \times 10^{-6}\right) \\ \varphi & =\frac{13}{3} \pi \end{aligned} $$
$$ \begin{aligned} \text { Now, } \quad I(\varphi) & =I _{\max } \cos ^{2} \frac{\varphi}{2} \\ \therefore \quad I & =I _{\max } \cos ^{2} \frac{13 \pi}{6} \\ I & =\frac{3}{4} I _{\max } \end{aligned} $$
(c) At $O$, path difference is $\Delta x=\Delta x _2=\frac{\mu _g}{\mu _m}-1 t$
For maximum intensity at $O$
$$ \begin{aligned} \Delta x & =n \lambda \\ \therefore \quad \lambda & =\frac{\Delta x}{1}, \frac{\Delta x}{2}, \frac{\Delta x}{3} \ldots \text { and so on } \\ \Delta x & =\frac{1.5}{4 / 3}-1\left(10.4 \times 10^{-6} m\right) \\ & =\frac{1.5}{4 / 3}-1\left(10.4 \times 10^{3} nm\right) \end{aligned} $$
$\Delta x=1300 nm$
$\therefore \quad$ Maximum intensity will be corresponding to
$$ \begin{aligned} \lambda & =1300 nm, \frac{1300}{2} nm, \frac{1300}{3} nm, \frac{1300}{4} nm \ldots \\ & =1300 nm, 650 nm, 433.33 nm, 325 nm \ldots \end{aligned} $$
The wavelengths in the range 400 to $700 nm$ are $650 nm$ and $433.33 nm$.
