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Posted

how do you derive that the refractive index of water with respect to air is real depth of an object placed in water/apparent depth as see from air?is this an approximation?

Posted

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?

  • 2 weeks later...
Posted

this is the diagram.

 

sorry about the delay in reply

 

You'll need to upload the image, or link to one on the web.

Posted (edited)

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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