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  • An object is placed at a depth of 10 cm inside a lake. The radius of cone through which this object can be seen is (refractive index of water =<img alt="\frac{3}{2}" src="https://entrancecorner.onc

An object is placed at a depth of 10 cm inside a lake. The radius of cone through which this object can be seen is (refractive index of water =\frac{3}{2})

Option: 1

7.95 cm 


Option: 2

8.95 cm 


Option: 3

9.95 cm


Option: 4

6.95 cm


Answers (1)

best_answer

As we learn

The radius of Circle of illuminance -

r=\frac{h}{\sqrt{\mu ^{2}-1}}

- wherein

 

\Rightarrow h=10 cm\Rightarrow r=\frac{10 cm}{\sqrt{\frac{9}{4}-1}}=\frac{10cm*2}{\sqrt{5}}

\mu =\frac{3}{2}

r = 8.95 cm

Posted by

vinayak

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