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(D) Let $PO = QO = x$. According to the sign convention,the object distance $u = -x$ and the image distance $v = +x$ (since the image is formed on the

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$6R$

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(D) Let $PO = QO = x$. According to the sign convention,the object distance $u = -x$ and the image distance $v = +x$ (since the image is formed on the other side of the refracting surface). The formula for refraction at a spherical surface is: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$ Here,$n_1 = 1$ (air) and $n_2 = 1.4$ (glass). Substituting the values: $\frac{1.4}{x} - \frac{1}{-x} = \frac{1.4 - 1}{R}$ $\frac{1.4}{x} + \frac{1}{x} = \frac{0.4}{R}$ $\frac{2.4}{x} = \frac{0.4}{R}$ $x = \frac{2.4}{0.4} R$ $x = 6R$ Therefore,the distance $PO = 6R$.

$A$ 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:

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