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Q 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?

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Given:

Actual height of the tank,h = 12.5 cm 

The apparent height of the tank, h' =  9.4 cm 

let the refractive index of the water be \mu

 \mu = \frac{h}{h'} =\frac{12.5}{9.4} = 1.33 (approx)

so the refractive index of water is approximately 1.33.

Now, when water is replaced with a liquid having  \mu = 1.63

\mu = \frac{h}{h'} =\frac{12.5}{h'_{new}} = 1.63

h'_{new}= \frac{12.5}{1.63}= 7.67 cm

Hence the new apparent height of the needle is 7.67 cm.

Total distance we have to move in a microscope = 9.4 - 7.67  = 1.73 cm.

Since the new apparent height is lesser than the previous apparent height we have to move UP the microscope in order to focus the needle.

Posted by

Pankaj Sanodiya

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