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# A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The center of curvature is in glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ. The distance PO is equal to R3R2R5R

A
5R
B
R
C
3R
D
2R
Solution
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#### Let us assumeOP = PQ = a (say)According to sign conventions,Object distance (u) = -xImage distance (v) = +xWe know the formulaμ2v - μ1u = μ2 - μ1RSubstituting the values in the above formula,1.5x - 1 - x = 1.5 - 1R1.5x + 1x = 0.5 R2.5x = 0.5 Rx = 5RThe distance PO is equal to 5R.

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