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Question
A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polystyrene are 1.58 and 1.55, respectively, for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen at a distance one metre away. (a) What would be the fringe-width? (b) At what distance from the centre will the first maximum be located?
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Solution
Given:-
The thickness of the strips = \[t_1 = t_2 = t = 0 . 5 mm = 0 . 5 \times {10}^{- 3} m\]
Separation between the two slits,
\[d = 0 . 12 cm = 12 \times {10}^{- 4} m\]
The refractive index of mica, μm = 1.58 and of polystyrene, μp = 1.58
Wavelength of the light,
\[\lambda = 590 nm = 590 \times {10}^{- 9} m,\]
Distance between screen and slit, D = 1 m
(a)
We know that fringe width is given by
\[\beta = \frac{\lambda D}{d}\]
\[\Rightarrow \beta = \frac{590 \times {10}^{- 9} \times 1}{12 \times {10}^{- 4}}\]
\[= 4 . 9 \times {10}^{- 4} m\]
(b) When both the mica and polystyrene strips are fitted before the slits, the optical path changes by
\[∆ x = \left( \mu_m - 1 \right) t - \left( \mu_p - 1 \right) t\]
\[= \left( \mu_m - \mu_p \right) t\]
\[ = \left( 1 . 58 - 1 . 55 \right) \times \left( 0 . 5 \right) \left( {10}^{- 3} \right)\]
\[ = \left( 0 . 015 \right) \times {10}^{- 3} m\]
∴ Number of fringes shifted, n = \[\frac{∆ x}{\lambda}\]
\[\Rightarrow n = \frac{0 . 015 \times {10}^{- 3}}{590 \times {10}^{- 9}} = 25 . 43\]
∴ 25 fringes and 0.43th of a fringe.
⇒ In which 13 bright fringes and 12 dark fringes and 0.43th of a dark fringe.
So, position of first maximum on both sides is given by
On one side,
\[x = \left( 0 . 43 \right) \times 4 . 91 \times {10}^{- 4}...........\left( \because \beta = 4 . 91 \times {10}^{- 4} m \right)\]
\[= 0 . 021 cm\]
On the other side,
\[x' = \left( 1 - 0 . 43 \right) \times 4 . 91 \times {10}^{- 4} \]
\[ = 0 . 028 cm\]
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