Optics 6 Question 53

55. 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: 55. (a) $0.63 mm \quad$ (b) $1.579 \mu m$

Solution:

  1. 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 $\frac{\lambda^{\prime}}{2}$

$$ \begin{aligned} & \therefore \quad \Delta x=\frac{\lambda^{\prime}}{2} \text { or } 0.15 t=\frac{4737}{2} \AA \\ & \therefore \quad t=15790 \AA \text { or } t=1.579 \mu m \end{aligned} $$



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