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Q.A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of tank is measured by a microscope to be 9.4 cm. What is the refractive index of water ? If water is replaced by 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 ?

Haryana BsehBSEH Intermediate Board 2025Subjective· 3mImportance★★★★★
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The refractive index is the real depth divided by the apparent depth (n ≈ 1.33 for water); with a denser liquid the apparent depth shrinks further, so the microscope must be lowered by about 1.73 cm to refocus.

Given: real depth of water h=12.5h = 12.5 cm, apparent depth h′=9.4h' = 9.4 cm.

Part 1 — Refractive index of water:

nwater=real depthapparent depth=12.59.4≈1.33n_{water} = \frac{\text{real depth}}{\text{apparent depth}} = \frac{12.5}{9.4} \approx 1.33

Part 2 — With the liquid of refractive index 1.63 (same real height 12.5 cm):

The new apparent depth is

hnew′=real depthnliquid=12.51.63≈7.67 cmh'_{new} = \frac{\text{real depth}}{n_{liquid}} = \frac{12.5}{1.63} \approx 7.67\ \text{cm}

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