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Hi everyone im new


Physicsfan

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I already did that. but after that i cannot go further.the image is formed at a depth lower than the apparent deapth because when light travels from a denser

to a rearer medium it bends away from normal and when the refracted ray is produced backward it meets with the ray travelling normal to the surface at a point above the actual position of object.how do i go further?

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  • 2 weeks later...

Ok. Now this is the image

 

 

Ok. this is the image.

 

This is how i understand it

 

the refractive index of water will be equal to sin r/sin i = (AO/sin i)/(AO/sin r)=K.

Now sin i=AB/OB, and sin r=AB/IB.

So numerator =AO/(AB/OB)=(AO*OB)/AB.

denominator=AO/(AB/IB)=(AO*IB)/AB.

 

So the refractive index is AO*OB/AO*IB=K.

=>OB/IB=K

 

Also, AO*OB/(AO/K)*KIB=K

So AO/(AO/K)=K (Is this correct?)

 

If correct, then AO/K=AI.

so refractive index=AO/AI.

 

 

If all this is correct, then can suggest me a simpler way of deriving this expression?

 

Thank u

refraction.psd

Edited by Physicsfan
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The real and apparent depths are going to be related by the tangent of the angles, with the common side AB. The only way this works for the relationship to index is to make an approximation for small angles, so you can assume the cosine terms are 1.

 

edit:

found a worked example with a picture

http://www.saburchill.com/physics/chapters3/0004.html

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