Optics Ques 172
- A double slit apparatus is immersed in a liquid of refractive index $1.33$. It has slit separation of $1 mm$ and distance between the plane of slits and screen is $1.33 $ $m$. The slits are illuminated by a parallel beam of light whose wavelength in air is $6300$ $ \AA$.
$(1996,3 M)$
(a) Calculate the fringe width.
(b) One of the slits of the apparatus is covered by a thin glass sheet of refractive index $1.53$. Find the smallest thickness of the sheet to bring the adjacent minimum as the axis.
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Answer:
Correct Answer: 172.(a) $0.63 $ $mm \quad$ (b) $1.579 $ $\mu m$
Solution:
Formula:
- Given, $\mu=1.33, d=1$ $ mm, D=1.33 $ $m$,
$ \lambda=6300$ $ \AA $
(a) Wavelength of light in the given liquid :
$ \begin{aligned} \lambda^{\prime} & =\frac{\lambda}{\mu}=\frac{6300}{1.33} \AA \approx 4737 \AA \\ & =4737 \times 10^{-10} m \end{aligned} $
$\therefore$ Fringe width, $\omega=\frac{\lambda^{\prime} D}{d}$
$ \begin{aligned} & \omega=\frac{\left(4737 \times 10^{-10} m\right)(1.33 m)}{\left(1 \times 10^{-3} m\right)}=6.3 \times 10^{-4} m \\ & \omega=0.63 mm \end{aligned} $
(b) Let $t$ be the thickness of the glass slab. Path difference due to glass slab at centre $O$.
$ \Delta x=(\frac{\mu _{\text {glass }}}{\mu _{\text {liquid }}}-1) t=(\frac{1.53}{1.33}-1) t $
$ \text { or } \Delta x=0.15 $ $t $
Now, for the intensity to be minimum at $s O$, this path difference should be equal to $\frac{\lambda^{\prime}}{2}$
$ \begin{aligned} & \therefore \quad \Delta x=\frac{\lambda^{\prime}}{2} \text { or } 0.15 t=\frac{4737}{2} \AA \\ & \therefore \quad t=15790 \AA \text { or } t=1.579 \mu m \end{aligned} $
BETA
