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Cool goIITian

Joined:	6 Oct 2011
Post:	35

6 Feb 2012 20:24:07 IST

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a beaker contains water and benzene of ref. indices 4/3 and 3/2 respectively. If their de

Physics

how do i calculate that?

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kavya bhat

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Joined: 6 Oct 2011

Posts: 35

6 Feb 2012 20:30:52 IST

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*if their depths are 12 cm each, the apparent depth of the bottom of the container is?

shivansh shrivastava

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Joined: 19 Jan 2012

Posts: 191

6 Feb 2012 22:44:16 IST

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I gues,u shud add two seperate apparent depths,by findin with formula ref.index=real.d/apparent.d and puttin real.d=12cm. . .im n0t sure bou that formula

Dipanshu Jaiswal

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Joined: 7 Feb 2012

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7 Feb 2012 00:59:53 IST

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app depth will be 3L/4 nd 2L/3 rispectively

kavya bhat

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Joined: 6 Oct 2011

Posts: 35

7 Feb 2012 17:11:11 IST

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i got the answer, u calculate the apparent depth of bottom using app depth= real depth/ ref. index , since water is denser and then calculate the depth of the benzene water interface using the same formula with ref. index of benzene. and then add the two, the answer comes to be 12/4/3=9 + 12/3/2=8 and so, 9+8= 17

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