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Question

A spherical convex surface of radius of curvature R separates air (μa=1) from glass(aμg=1.5) .The centre of curvature is in the glass is O. 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

A
5R
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B
3R
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C
2R
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D
1.5 R
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Solution

The correct option is A 5R
Using the relation μ2vμ1u=μ2μ1R
where u and v is the object distance and image distance, respectively.
Given : μ1=1 and μ2=1.5
Let the distance of object from point O be x.
u=x and v=x
1.5xμ1(x)=1.51R
32x+1x=12R

3+22x=12R
x=5R

54241_5040_ans_fbb7b28791c943fc8727749be5a9ccb8.png

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