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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Jul 2007 13:36:32 IST
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Consider the situation.The bottom of the pot (heigth is H) is a reflecting plane mirror.A fish is floting on the water at a height of H/2 from the mirror and your eye is H distance away from the pot.At what distance from itself will the eye (yours)see the image(s) of the fish. And do rate me plzzzzzzzzzzzzzzzz
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6 Jul 2007 13:44:33 IST
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is it 9/8H
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6 Jul 2007 15:01:33 IST
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No it is not the answer
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14 Jul 2007 22:19:33 IST
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The answer is 2H. Do reply to confirm it.
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14 Jul 2007 23:23:30 IST
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answer is 17r/8...himanshu was almost rite,but he 4got 2 add r to the final answer...
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14 Jul 2007 23:35:33 IST
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Could you just explain how you got that expression cool shetty
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15 Jul 2007 00:09:30 IST
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a little elaboration plz
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it is not important where u stand, but in which direction u are moving |
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15 Jul 2007 00:58:42 IST
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is the answer 3H/2.
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RATE ME FOR MY EFFORTS....... |
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16 Jul 2007 19:26:29 IST
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I am sorry.I accutally forgot something.Consider the refrative index of the water as 'mu'.And the answers are wrong.Keep tring you migth get it(you can say that I am quizzing you).
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16 Jul 2007 19:38:11 IST
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Then the answer is 3 H/2
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16 Jul 2007 23:06:57 IST
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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.
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17 Jul 2007 00:28:15 IST
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am in full agreement wid elessar_iitkgp; its written floating 'on' water, missed it in casualness, thought dat the pot is full; hats off 2 u dude
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