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A glass filled with water upto 12.5cm. The apparent of an object lying at the bottom of the glass is measured by a microscope to be 9.4m. Calculate 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 the object again?

Rai , 8 Years ago
Grade 12
anser 1 Answers
Orooba Asim

Last Activity: 8 Years ago

Actual depth, h1 = 12.5 cm
Apparent depth, h2= 9.4 cm
Refractive index n1= h1/h2= 1.33  Therefore nH2O = 1.33
Refractive index of the other liquid is n’ = 1.63
(Actual depth) h1= 12.5 bc it’s the same glass.
Apparent depth h’ = h1/ n’ = 7.7 (approx.) and h’ 2 and thus the microscope should be moved upwards by a distance of h2 - h’ = 1.7 cm 
 

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