A fish in an aquarium, 30 em deep in water can see a light bulb kept 50 em above the surface of water. The fish can also see the image of this bulb in the reflecting bottom surface of the aquarium. Total depth of water is 60 ern. Then the. apparent distance between the two images seen by the fish is ( $μ_{w}$ = 4/3).

140 m

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\frac{760}{3} m

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\frac{280}{3} c m

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\frac{380}{3} c m

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Solution

The correct option is A $\frac{760}{3} m$
Apparent distance ofthe bulb from the fish
$d_{1} = 50 μ + 30$
apparent distance of the image
$d_{2} = 50 μ + 60 + 30$
$∴ d_{1} + d_{2} = 100 μ + 120$

$= \frac{400}{3} + 120 = \frac{760}{3} c m$

$= 253.3 c m$

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The correct option is A 7603mApparent distance ofthe bulb from the fishd1=50μ+30apparent distance of the imaged2=50μ+60+30∴d1+d2=100μ+120=4003+120=7603cm=253.3cm

The correct option is A $\frac{760}{3} m$
Apparent distance ofthe bulb from the fish
$d_{1} = 50 μ + 30$
apparent distance of the image
$d_{2} = 50 μ + 60 + 30$
$∴ d_{1} + d_{2} = 100 μ + 120$

$= \frac{400}{3} + 120 = \frac{760}{3} c m$

$= 253.3 c m$