A spherical surface of radius of curvature R | vedclass

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

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$5$

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(A) When a ray of light travels from $\mu_{1}$ to $\mu_{2}$ after refraction at a single curved surface,the formula is:$\frac{\mu_{2}}{v} - \frac{\mu_{1}}{u} = \frac{\mu_{2} - \mu_{1}}{R}$As per the sign convention:$u = -x$ (object is in air,distance $PO = x$)$v = +x$ (image is real and in glass,distance $OQ = x$)$R$ is positive as the center of curvature is in the glass.Given $\mu_{1} = 1.0$ and $\mu_{2} = 1.5$:$\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 = 5R$Therefore,the distance $PO = 5R$.

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