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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Physics
Chapter
Wave Optics

BITSAT 2022 · Physics · Wave Optics

In a young's double slit arrangement frings are produced using...

In a young's double slit arrangement frings are produced using light of wavelength 4000A˚4000 Å. One slit is covered by a thin plate of glass of refractive index 1.41.4 and the other with another glass plate of same thickness but of refractive index 1.7. By doing so the central bright shifts to original sixth fringe from centre. Thickness of glass plate is

Options

  1. A2μm2 \mu \mathrm{m}
  2. B8μm8 \mu \mathrm{m}
  3. C11μm11 \mu \mathrm{m}
  4. D16μm16 \mu \mathrm{m}

Correct answer

B. 8μm8 \mu \mathrm{m}

Step-by-step solution

As, shift Δx=βλ(μ−1)t\Delta x=\frac{\beta}{\lambda}(\mu-1) t Shift due to one plate Δx1=βλ(μ1−1)t\Delta x_1=\frac{\beta}{\lambda}\left(\mu_1-1\right) t Shift due to another path Δx2=βλ(μ2−1)t\Delta x_2=\frac{\beta}{\lambda}\left(\mu_2-1\right) t Net shift Δx=Δx2−Δx1=βλ(μ2−μ1)t\Delta x=\Delta x_2-\Delta x_1=\frac{\beta}{\lambda}\left(\mu_2-\mu_1\right) t Also, it is given that Δx=6β\Delta x=6 \beta Hence, 6β=βλ(μ1−μ2)t⇒t=6λ(μ1−μ2)=6×4000(1.7−1.4)−80000A˚⇒t=8×10−6=8μm\begin{aligned} \text{ Hence, }6 \beta=\frac{\beta}{\lambda}\left(\mu_1-\mu_2\right) t \\ & \Rightarrow \quad t=\frac{6 \lambda}{\left(\mu_1-\mu_2\right)}=\frac{6 \times 4000}{(1.7-1.4)}-80000 Å \\ & \Rightarrow \quad t=8 \times10^{-6}=8 \mu \mathrm{m} \end{aligned}$$
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