Ncert Exercises & Solutions

9.3 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 same height, by what distance would the microscope have to be moved to focus on the needle again?

Hint: $Apparent~depth=\frac{Real~depth}{\mu}$
Step 1: Find the refractive index of water.
The actual depth of the needle, $h_{1}$ =12.5cm
The apparent depth of the needle, $h_{2}$ =9.4 cm
Refractive index of water, $μ = \frac{a c t u a l d e p t h}{a p p e r e n t d e p t h} = \frac{h_{1}}{h_{2}} = \frac{12.5}{9.4} = 1.33$
Step 2: Find the new apparent depth of the needle.
When water is replaced by a liquid of refractive index , $μ^{'} =$ 1.63
The new apparent depth of the needle, $h = \frac{h_{1}}{μ^{'}} = \frac{12.5}{1.63} = 7.67 c m < h_{2}$
Step 3: Find the neddle shift.
To focus the needle again, the microscope should be moved up by a distance = 9.4 - 7.67 = 1.73 cm

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