JEE MAIN - Physics (2005 - No. 5)
A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is $${4 \over 3}$$ and the fish is $$12$$ $$cm$$ below the surface, the radius of this circle in $$cm$$ is
$${{36} \over {\sqrt 7 }}$$
$${36\sqrt 7 }$$
$${4\sqrt 5 }$$
$${36\sqrt 5 }$$
Explanation
$$\sin {\theta _c} = {1 \over \mu } = {3 \over 4}$$
or $$\tan {\theta _c} = {3 \over {\sqrt {16 - 9} }} = {3 \over {\sqrt 7 }} = {R \over {12}}$$
$$ \Rightarrow R = {{36} \over {\sqrt 7 }}\,cm$$
or $$\tan {\theta _c} = {3 \over {\sqrt {16 - 9} }} = {3 \over {\sqrt 7 }} = {R \over {12}}$$
$$ \Rightarrow R = {{36} \over {\sqrt 7 }}\,cm$$
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