As per the wording of the question "A fish is floating on the water" it seems that the pot is half full Then the man sees the fish in front of his eyes at a distance H/2 Now consider a ray traveling towards the bottom of the pot from the fish. Its image is formed at a distance of H/2 below the bottom of pot. This acts the object for the refraction of the ray at the water-air interface. Distance of this image from the water air interface is H. So distance of the image of this virtual object from this interface = H/. So distance of the image from the man's eyes = H/2 + H/. So there is only one image and the other one is the fish itself.
The last line of the question is a bit confusing. Is the man's eye H distance away from the top of the pot or the bottom of the pot. Please clarify. I have assumed it to be H distance from the bottom of the pot.
If its at a distance H from the top of the pot, Distance of the fish = 3H/2 And distance of the image = 3 H/2 + H/
However if the pot is full to the top, then and the man is seeing from H distance from the top, Now consider a ray traveling towards the top of the pot from the fish. It is at a distance of H/2 from the water-air interface. Its image is formed H/2 from the water surface. Then the distance of this image from the eye is H + H/2 Now consider a ray traveling towards the bottom of the pot from the fish. Its image is formed at a distance of H/2 below the bottom of pot. This acts the object for the refraction of the ray at the water-air interface. Distance of this image from the water air interface is 3H/2. So distance of the image of this virtual object from this interface = 3H/2. So distance of the second image fromfrom the man's eyes = H + 3H/2.
Moderation Team
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