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I've found the following question on IsaacPhysics.org . I've been working on it for a couple of hours but I cannot figure it out.

A looking glass is used to read an ancient scroll. This lens has a hemispherical shape with radius R and its flat surface is placed on top of the scroll being examined. The lens is made of glass of refractive index n. The reader is a distance h from the scroll along the axis of symmetry.

What is the minimum value of h for which the entire circular area of the scroll directly under the lens is visible to the observer if $R=3.70\ \rm cm$ and $n=1.30$?

enter image description here

enter image description here

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closed as off-topic by ZeroTheHero, user191954, John Rennie, Kyle Kanos, Jon Custer Sep 13 '18 at 12:52

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  • $\begingroup$ I have tried to workout at what angle there would be total internal reflection from a light ray coming from the edge of the scroll. My reasoning was that at a greater angle than that no light would pass and the scroll would thus not be entirely visible. My difficulty is that this would give a maximal height above which the scroll would no longer visible, eventhough the question specifically asks for a minimal value. $\endgroup$ – user206444 Sep 12 '18 at 17:58
  • $\begingroup$ I've added an image to the question to further elaborate on my comment. $\endgroup$ – user206444 Sep 12 '18 at 18:12
  • $\begingroup$ Excellent work. You have solved the problem and shown that the question is incorrectly worded. Minimum should be maximum. Websites and textbooks can contain errors. Have more trust in your own abilities. $\endgroup$ – sammy gerbil Sep 12 '18 at 18:39
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I have tried to workout at what angle there would be total internal reflection. My reasoning was that at a greater angle than that no light would pass and thus the scroll not entirely visible.

I think your direction is correct. Here is a diagram that might help you figure that out.

enter image description here

Red lines represents the light path from points A and B of the scroll to the observer at point E.

If point E is moved down, you may find that, for the Snell's law to hold for the BDE light path, point D would have to move down the circle and line DE would become closer to the tangential line, which would indicate that we are approaching the point of the total internal reflection.

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