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(A) When a light ray travels from $\mu_{1}$ to $\mu_{2}$ after refraction at a single spherical surface,the formula is given by:$\frac{\mu_{2}}{v} - \

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$5$

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(A) When a light ray travels from $\mu_{1}$ to $\mu_{2}$ after refraction at a single spherical surface,the formula is given by:$\frac{\mu_{2}}{v} - \frac{\mu_{1}}{u} = \frac{\mu_{2} - \mu_{1}}{R}$Given that the object is in air $(\mu_{1} = 1.0)$ and the image is in glass $(\mu_{2} = 1.5)$.According to the sign convention,the object distance $u = -PO = -x$ and the image distance $v = +OQ = +x$ (since $PO = OQ = x$). The radius of curvature $R$ is positive as the center of curvature lies in the glass.Substituting these values into the formula:$\frac{1.5}{x} - \frac{1}{-x} = \frac{1.5 - 1.0}{R}$$\frac{1.5}{x} + \frac{1}{x} = \frac{0.5}{R}$$\frac{2.5}{x} = \frac{0.5}{R}$$x = \frac{2.5}{0.5} R = 5 R$Therefore,the distance $PO$ is $5 R$.

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