MCQ
In a Young's double slit experiment with light of wavelength $\lambda ,$ fringe pattern on the screen has fringe width $\beta .$ When two thin transparent glass (refractive index $\mu$ ) plates of thickness $t_1$ and $t_2\,\,(t_1 > t_2)$ are placed in the path of the two beams respectively, the fringe pattern will shift by a distance
  • A
    $\frac{{\beta (\mu  - 1)}}{\lambda }\left( {\frac{{{t_1}}}{{{t_2}}}} \right)$
  • B
    $\frac{{\mu \beta }}{\lambda }\frac{{{t_1}}}{{{t_2}}}$
  • $\frac{{\beta \left( {\mu  - 1} \right)}}{\lambda }({t_1} - {t_2})$
  • D
    $\frac{{\left( {\mu  - 1} \right)\lambda }}{\beta }({t_1} + {t_2})$

Answer

Correct option: C.
$\frac{{\beta \left( {\mu  - 1} \right)}}{\lambda }({t_1} - {t_2})$
c
Shift $=\frac{\beta(\mu-1)}{\lambda} t_{1}-\frac{\beta(\mu-1)}{\lambda} t_{2}$

$ = \frac{{\beta (\mu  - 1)}}{\lambda }\left( {{t_1} - {t_2}} \right)$

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