A fish under water at a depth of 20 cm can see the outer atmosphere through an aperture of radius of (critical angle of water is $45^{\circ}$ )

5 c m

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20 c m

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10 c m

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\frac{20}{\sqrt{3}} c m

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Solution

The correct option is B $20 c m$
So, $tan (\frac{π}{2} - θ_{c}) = \frac{h}{R}$
$tan (90^{\circ} - 45^{\circ}) = \frac{20}{R}$ cm
$R = \frac{20 c m}{tan 45^{\circ}}$
$= \frac{20}{1}$ cm $(∵ tan 45^{\circ} = 1)$

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A fish under water at a depth of 20 cm can see the outer atmosphere through an aperture of radius of (critical angle of water is 45∘)

The correct option is B 20 cmSo, tan(π2−θc)=hRtan(90∘−45∘)=20R cmR=20cmtan45∘=201 cm (∵tan45∘=1)

A fish under water at a depth of 20 cm can see the outer atmosphere through an aperture of radius of (critical angle of water is $45^{\circ}$ )

The correct option is B $20 c m$
So, $tan (\frac{π}{2} - θ_{c}) = \frac{h}{R}$
$tan (90^{\circ} - 45^{\circ}) = \frac{20}{R}$ cm
$R = \frac{20 c m}{tan 45^{\circ}}$
$= \frac{20}{1}$ cm $(∵ tan 45^{\circ} = 1)$