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Optics  Refraction and a transparent sphere 
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#1
Feb1613, 09:33 AM

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1. The problem statement, all variables and given/known data
A spider is hanging by means of its own silk thread directly above a transparent fixed sphere of radius R=20 cm as shown in the figure. The refractive index of the material of the sphere is equal to ##\sqrt{2}## and the height of the spider from the centre of the sphere is 2R. An insect, initially sitting at the bottom, starts crawling on the sphere along a vertical circular path with the constant speed of ##\frac{\pi}{4}## cm/s. For how long time the insect will be invisible for the spider, assume that it crawls once round the vertical circle. (see attachment) 2. Relevant equations 3. The attempt at a solution Honestly, I have no idea. I don't have a clue about which equations I have to use here. Any help is appreciated. Thanks! 


#2
Feb1613, 10:17 AM

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#3
Feb1613, 09:27 PM

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#4
Feb1613, 09:41 PM

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Optics  Refraction and a transparent sphere
I actually found this easier to do by tracing the ray from the spider, back to the insect. 


#5
Feb1713, 10:29 PM

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If a ray from the spider is tangent to the sphere, then by Snell's law, it deviates towards the normal by pi/4. (see attachment) I am facing problems with geometry. Look at the triangle where the angles come out to be 150 and 45 degrees, the sum goes over 180 degrees! I can't find where I went wrong. EDIT: Woops, please ignore this reply. I think I have found out my mistake. One more question, would the rays meet, after refraction, at the point where insect was initially placed? 


#6
Feb1713, 11:01 PM

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The angle you have labelled as 60° is 120° . 


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Feb1713, 11:02 PM

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#8
Feb1713, 11:09 PM

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#9
Feb1813, 12:03 AM

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I am still not getting the right answer.
(see attachment) The angular displacement of the insect when it is invisible to the spider is ##2\cdot \frac{13\pi}{24}##. The time taken is ##t=\frac{\theta}{\omega}## where ##\omega## is the angular velocity of insect. Substituting the values, ##t=\frac{260}{3} sec## which is wrong. :( 


#10
Feb1813, 01:01 AM

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Your drawings are misleading.
The refracted ray makes an angle of 45° (π/2) with respect to the normal . In both drawings, the refracted ray appears to be drawn at an angle of 45° with respect to the horizontal. 


#11
Feb1813, 01:04 AM

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#12
Feb1813, 01:14 AM

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The bug is visible along the red arc. The refractive index is 1/√2 so the angle of refraction is 45°. The radii make 90°angle between the point of incidence and point where the refracted ray arrives to the circle again.
ehild 


#13
Feb1813, 01:23 AM

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Thank you both for your inputs, I have got the right answer!



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Feb1813, 01:24 AM

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Feb1813, 01:27 AM

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#16
Feb1813, 01:29 AM

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#17
Feb1813, 01:32 AM

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Yes but see my previous diagrams, they are completely wrong. 


#18
Feb1813, 01:36 AM

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Yes, the upper central angle was wrong, and also, you did not draw the refracted light according to the angle you got.
ehild. 


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