Q140 / 150 · MHT CET 2022 (07 Aug Shift 1) · Physics

On placing a thin film of mica of thickness $12 \times 10^{-5} \mathrm{~cm}$ in the path of one of the interfering waves in Young's double slit experiment using monochromatic light, the fringe pattern shifts through a distance equal to width of a bright fringe. If wavelength used is $\lambda=6 \times 10^{-5} \mathrm{~cm}$, the refractive index of mica is

  • A
    $1.4$
  • B
    $1.1$
  • C
    $1.3$
  • Correct answer
    D
    $1.5$

Solution

Path difference due to the mica film: $(\mu-1) \mathrm{t}$ The shift in the pattern be $\Delta$ yand the corresponding path difference can be written as: $\Delta x=\frac{\Delta y D}{d}=(\mu-1) t$ $\therefore \Delta y=\frac{d}{D}(\mu-1) t$ Fringe width is $\beta=\frac{\lambda D}{d}$, so, $\Delta y=\frac{\beta}{\lambda}(\mu-1) t$ If the shift is equal to the fringe width $\beta$ $\begin{aligned} & \beta=\frac{\beta}{\lambda}(\mu-1) t \\ & \Rightarrow(\mu-1) t=\lambda \\ & \therefore \mu=\frac{\lambda}{t}+1=\left(\frac{6 \times 10^{-5}}{12 \times 10^{-5}}\right)+1=1.5\end{aligned}$

Hence, the correct answer is option: 'D'.

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