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Karnataka Board PUCPUC Science Class 11

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. - Physics

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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?

Sum

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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Chapter 17: Light Waves - Exercise [Page 381]

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HC Verma Concepts of Physics Vol. 1 [English] Class 11 and 12
Chapter 17 Light Waves
Exercise | Q 16 | Page 381

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