MHT CET · Physics · Wave Optics
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 \(1.3\)
- D \(1.5\)
Answer & Solution
Correct Answer
(D) \(1.5\)
Step-by-step Solution
Detailed explanation
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 \(\beta\)
\(\begin{aligned} & \beta=\frac{\beta}{\lambda}(\mu-1) t \\ & \Rightarrow(\mu-1) t=\lambda \\ & \therefore \mu=\frac{\lambda}{t}+1=\left(\frac{6 \times 10^{-5}}{12 \times 10^{-5}}\right)+1=1.5\end{aligned}\)
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 \(\beta\)
\(\begin{aligned} & \beta=\frac{\beta}{\lambda}(\mu-1) t \\ & \Rightarrow(\mu-1) t=\lambda \\ & \therefore \mu=\frac{\lambda}{t}+1=\left(\frac{6 \times 10^{-5}}{12 \times 10^{-5}}\right)+1=1.5\end{aligned}\)
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