$A$ 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 (in $R$)

  • A
    $5$
  • B
    $3$
  • C
    $2$
  • D
    $1.5$

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$A$ spherical surface of radius of curvature $R$ separates air from glass (refractive index $= 1.5$). The centre of curvature is in the glass medium. $A$ point object $O$ is placed in air on the optic axis of the surface,so that its real image is formed at $I$ inside the glass. The line $OI$ intersects the spherical surface at $P$ and $PO = PI$. The distance $PO$ is equal to: (in $R$)

$A$ point object is kept at $P$ in front of a glass sphere of radius $R$. Its image is formed at $Q$ such that $PO = QO$. The refractive index of the material of the glass sphere is $1.4$. The distance $PO$ is equal to:

$A$ curved surface of radius $R$ separates two media of refractive indices $\mu_1$ and $\mu_2$ as shown in figures $A$ and $B$. Choose the correct statement$(s)$ related to the virtual image formed by object $O$ placed at a distance $x$,as shown in figure $A$.

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