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Q.A tank is filled with water to a height of 12.5 cm. The apparent depth of the needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 upto the same height, by what distance would the microscope have to be moved to focus on the needle again? OR Find the position of the image formed by the lens combination given in the figure.

Nagaland NbseNagaland Board of School Education 2024Subjective· 3mImportance★★★★★
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Refractive index = real depth ÷ apparent depth; using this both to find water's index and, for the new liquid, the new (smaller) apparent depth, whose difference from the old one is the distance the microscope must be moved.

Refractive index of water: For near-normal viewing, the simple relation for apparent depth is

μ=real depthapparent depth\mu = \frac{\text{real depth}}{\text{apparent depth}}

Given real depth = 12.5 cm, apparent depth = 9.4 cm:

μwater=12.59.4≈1.330\mu_{water} = \frac{12.5}{9.4} \approx 1.330

Distance to move the microscope, for the new liquid: With the tank filled to the SAME real depth (12.5 cm) but now with a liquid of refractive index μ′=1.63\mu'=1.63, the new apparent depth is

apparent depth′=real depthμ′=12.51.63≈7.67 cm\text{apparent depth}' = \frac{\text{real depth}}{\mu'} = \frac{12.5}{1.63} \approx 7.67\,\text{cm}

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