Question:
A tank is filled with water to a height of 12.5 cm. The apparent depth of a 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 up to the sameheight, by what distance would the microscope have to be moved to focus on the needle again?
Answer:
Actual depth of the needle in water (h1) = 12.5 cm
Apparent depth of the needle in water (h2) = 9.4 cm
Refractive index of water = μ
The value of μ can be obtained
μ = h1/ h2 = 12.5/9.4 = 1.33
Hence, the refractive index of water is about 1.33.
Water is replaced by a liquid of refractive index (μ1) = 1.63
The actual depth of the needle remains the same, but its apparent depth changes. Let y be the new apparent depth of the needle.
Hence,
μ1 = h1/y
y = h1/μ1
= 12.5/1.63 = 7.67 cm
Distance by which the microscope should be moved up = 9.4 – 7.67 = 1.73 cm
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