Geometrical OpticsMedium
Question
A fish in an aquarium, 30 cm deep in water can see a light bulb kept 50 cm above thesurface 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 cm. then the appartent distance be tween the two images the apparent distance between the two images seen by the fish is (μw = 4/3)
Options
A.140 cm
B.
cm
cm C.
cm
cm D.
cm
cm Solution
Appartent distance of the bulb of from the fish d1 = 50μ + 30
apparemt distance of the image
d2 = 50μ + 60 + 30
∴ d1 + d2 = 100μ + 120
=
+ 120 = 
cm= 253.3cm
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