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Question

A spherical surface of radius of curvature R, separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the 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 :

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

The correct option is A 5R
Given, PO=OQ

u=v

We know

μ2vμ1u=μ2μ1R

μ2vμ1u=μ2μ1R

μ2+μ1V=μ2μ1R

1.5+1v=1.51R

2.5v=0.5R

v=2.5R0.5

v=5R

OP=OQ=u=v=5R

1428920_1196653_ans_4a232a710ad4400c8bd63be8780ff39f.png

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