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  • Interference fringes were produced by Young's double slit method, the wavelength of light used being <img alt="6000 \mathrm{~A}^{0}" src="https://entrancecorner.oncodecogs.com/gif.latex?6000%20

Interference fringes were produced by Young's double slit method, the wavelength of light used being 6000 \mathrm{~A}^{0}. The separation between the two slits is 2 \mathrm{~mm}.The distance between the slits and screen is 10 \mathrm{~cm}.When a transparent plate of thickness 0.5 \mathrm{~mm} is placed over one of the slits, the fringe pattern is displaced by 5 \mathrm{~mm}. The refractive index of the material of the plate is:

Option: 1

1.20


Option: 2

1.50


Option: 3

1.25


Option: 4

1.80


Answers (1)

best_answer

Here 2 \mathrm{~d}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}, \quad D=10 \mathrm{~cm}=0.10 \mathrm{~m},
\mathrm{t}=0.5 \mathrm{~mm}=0.5 \times 10^{-3} \mathrm{~m}, \Delta \mathrm{x}=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m}, \lambda=6 \times 10^{-7} \mathrm{~m}

\mathrm{As \, x_{0}=\frac{D}{2 d}(\mu-1) t}

\mathrm{\therefore \quad \mu-1=\frac{\Delta x(2 d)}{D t}=\frac{5 \times 10^{-3} \times 2 \times 10^{-3}}{0.10 \times 0.5 \times 10^{-3}}=0.2 \quad \text { or } \quad \mu=1+0.2=1.2}

Posted by

Sanket Gandhi

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