Physicsfan Posted April 23, 2010 Share Posted April 23, 2010 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? Link to comment Share on other sites More sharing options...
swansont Posted April 23, 2010 Share Posted April 23, 2010 Try drawing a picture and applying Snell's law. Link to comment Share on other sites More sharing options...
Physicsfan Posted April 25, 2010 Author Share Posted April 25, 2010 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? Link to comment Share on other sites More sharing options...
swansont Posted April 25, 2010 Share Posted April 25, 2010 You should be able to draw some triangles to relate real depth and apparent depth. Link to comment Share on other sites More sharing options...
Physicsfan Posted May 10, 2010 Author Share Posted May 10, 2010 this is the diagram. sorry about the delay in reply Link to comment Share on other sites More sharing options...
swansont Posted May 10, 2010 Share Posted May 10, 2010 this is the diagram. sorry about the delay in reply You'll need to upload the image, or link to one on the web. Link to comment Share on other sites More sharing options...
Physicsfan Posted May 12, 2010 Author Share Posted May 12, 2010 (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 May 12, 2010 by Physicsfan Link to comment Share on other sites More sharing options...
Physicsfan Posted May 13, 2010 Author Share Posted May 13, 2010 the AO/K=AI part is the problem. that part is where i get stuck. Link to comment Share on other sites More sharing options...
swansont Posted May 13, 2010 Share Posted May 13, 2010 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 Link to comment Share on other sites More sharing options...
Physicsfan Posted May 13, 2010 Author Share Posted May 13, 2010 so it is an approximation then! nice web page thank u! Link to comment Share on other sites More sharing options...
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