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?

Try drawing a picture and applying Snell's law.

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?

You should be able to draw some triangles to relate real depth and apparent depth.

this is the diagram.

 

sorry about the delay in reply

this is the diagram.

 

sorry about the delay in reply

 

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

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 May 12, 201015 yr by Physicsfan

the AO/K=AI part is the problem.

that part is where i get stuck.

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

so it is an approximation then!

nice web page

 

thank u!