The 7th bright fringe in the Young’s double slit experiment using a light of wavelength 550 nm shifts to the central maxima after covering the two slits with two sheets of different refractive indices $n_1$ and $n_2$ but having the same thickness 6 µm. The value of $ |n_1 - n_2| $ is .............

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Step 1: Understand the shift in fringe pattern. In the Young’s double slit experiment, the position of the bright fringes is given by the equation: $$ y_m = \frac{m\lambda D}{d} $$ where $ y_m $ is the distance of the $ m $-th bright fringe from the central maximum, $ \lambda $ is the wavelength of the light, $ D $ is the distance between the screen and the slits, and $ d $ is the separation between the slits. When two sheets with different refractive indices are placed in front of the slits, the effective wavelength of light inside the sheets changes. The new wavelength $ \lambda_{\text{eff}} $ inside the medium is given by: $$ \lambda_{\text{eff}} = \frac{\lambda}{n} $$ where $ n $ is the refractive index of the medium. Step 2: Calculate the shift in the 7th bright fringe. The 7th bright fringe corresponds to $ m = 7 $. The shift in position due to the change in refractive index is given by the difference in the positions before and after the change in medium. Let $ \Delta y_7 $ be the shift in the 7th fringe: $$ \Delta y_7 = \frac{7\lambda}{d} \left( \frac{1}{n_1} - \frac{1}{n_2} \right) $$ Given that the 7th bright fringe shifts to the central maximum, we can set $ \Delta y_7 $ equal to the thickness of the sheets, which is 6 µm. Thus: $$ 6 \times 10^{-6} = \frac{7 \times 550 \times 10^{-9}}{d} \left( \frac{1}{n_1} - \frac{1}{n_2} \right) $$ Solving for $ |n_1 - n_2| $: $$ |n_1 - n_2| = \frac{6 \times 10^{-6} \times d}{7 \times 550 \times 10^{-9}} $$ Step 3: Conclusion. Thus, the value of $ |n_1 - n_2| $ is approximately 0.04 .

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Correct Answer: 0.63

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