Q143 / 160 · AP EAMCET 2022 (05 Jul Shift 1) · Physics
Young's double slit experiment is conducted with monochromatic light of wavelength $5000 Å$, with slit separation of $3 \mathrm{~mm}$ and observer at $20 \mathrm{~cm}$ away from the slits. If a $1 \mathrm{~mm}$ transparent plate is placed infront of one of the slits, the fringes shift by $6 \mathrm{~mm}$. The refractive index of the transparent plate is
- A$1.08$
- Correct answerB$1.09$
- C$1.1$
- D$1.2$
Solution
Given, wavelength, $\lambda=5000 Å=5 \times 10^{-7} \mathrm{~m}$
Slit separation, $d=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}$
Distance between slit and screen,
$D=20 \mathrm{~cm}=0.2 \mathrm{~m}$
Thickness of transparent plate $=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}$
Fringe shift, $\Delta x=6 \mathrm{~mm}=6 \times 10^{-3} \mathrm{~m}$
Fringe shift' is given by
$\Delta x=\frac{D}{d}(\mu-1) t$
$\Rightarrow \quad 6 \times 10^{-3}=\frac{0.2}{3 \times 10^{-3}}(\mu-1) \times 10^{-3}$
$\Rightarrow \frac{6 \times 3 \times 10^{-3}}{0.2}=\mu-1 \Rightarrow 9 \times 10^{-2}=\mu-1$
$\Rightarrow \quad \mu=1.09$
Hence, refractive index of transparent plate is 1.09 .
Hence, the correct answer is option: 'B'.