A pond of water is $5 m$ deep. The flame is held $2 m$ above the surface of water. A fish is at depth $4 m$ from water surface. The refractive index of water is $4 / 3$ . The apparent height of flame from the eyes of fish is

\frac{8}{3} m

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6 m

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5.5 m

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\frac{20}{3} m

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Solution

The correct option is D $\frac{20}{3} m$
Given : $h_{1} = 2$ m $h_{2} = 4$ m $n_{w} = 4 / 3$ $n_{1} = 1$

Apparent height of flame as seen by fish, $h_{o b s} = n_{w} [\frac{h_{1}}{n_{1}} + \frac{h_{2}}{n_{w}}]$

$∴$ $h_{o b s} = \frac{4}{3} [\frac{2}{1} + \frac{4}{4 / 3}] = \frac{20}{3}$ m

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A pond of water is 5m deep. The flame is held 2m above the surface of water. A fish is at depth 4m from water surface. The refractive index of water is 4/3. The apparent height of flame from the eyes of fish is

The correct option is D 203mGiven : h1=2 m h2=4 m nw=4/3 n1=1Apparent height of flame as seen by fish, hobs=nw[h1n1+h2nw]∴ hobs=43[21+44/3]=203 m

A pond of water is $5 m$ deep. The flame is held $2 m$ above the surface of water. A fish is at depth $4 m$ from water surface. The refractive index of water is $4 / 3$ . The apparent height of flame from the eyes of fish is

The correct option is D $\frac{20}{3} m$
Given : $h_{1} = 2$ m $h_{2} = 4$ m $n_{w} = 4 / 3$ $n_{1} = 1$

Apparent height of flame as seen by fish, $h_{o b s} = n_{w} [\frac{h_{1}}{n_{1}} + \frac{h_{2}}{n_{w}}]$

$∴$ $h_{o b s} = \frac{4}{3} [\frac{2}{1} + \frac{4}{4 / 3}] = \frac{20}{3}$ m