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Old Jun 25th 2009, 09:49 AM   #1
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refraction at plane surfaces

a glass sphere in the shape of a quarter - cylinder lies on ahorizontal table . a uniform , horizontal light beam falls on
its vertical plane surfaces, as shown in the figure . if the radius of the cylinder is R = 5 cm and the refractive index of the glass is n = 1.5 .

where , on the table beyond the cylinder , will a path of light be found??????
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Old Jun 25th 2009, 11:10 PM   #2
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The critical angle is given by sinC = 1/n.
The parallel beam of light making more than critical angle with the curved surface of the quarter cylinder, will not come out of the cylinder. The ray making the angle of incidence equal to critical angle will come out of the glass tangential to the cylindrical surface. If O is the center and PQ is the tangent, in the right angled triangle OPQ angle POQ is equal to C and cosC will give you the first image of the ray.
When the ray is very close to the base of the cylinder the image position is given by
n/u + 1/v = (n-1)/R. Here u = infinity, R = 5 cm. If S is the image of this ray, QS is the image on the table due to the parallel beam of light.
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