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JEE Main & Advanced Physics Ray Optics Question Bank Critical Thinking

  • question_answer
    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                             [IIT JEE 1998; DPMT 2000]

    A)            5 R 

    B)            3 R

    C)            2 R 

    D)            1.5 R

    Correct Answer: A

    Solution :

                       \[\frac{{{\mu }_{2}}}{v}-\frac{{{\mu }_{1}}}{u}=\frac{{{\mu }_{2}}-{{\mu }_{1}}}{R}\Rightarrow \frac{1.5}{+OQ}-\frac{1}{(-OP)}=\frac{(1.5-1)}{+R}\]            On putting \[OQ=OP,\,\,OP=5R\]


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