A spherical surface of radius of curvature R | vedclass

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(D) For refraction at a spherical surface,the formula is given by: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.Here,$n_1 = 1.0$ (air) and $n

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$5\, R$

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(D) For refraction at a spherical surface,the formula is given by: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.Here,$n_1 = 1.0$ (air) and $n_2 = 1.5$ (glass).The object is at $P$ in air,so $u = -PO$. Let $PO = x$,then $u = -x$.The image is real and formed at $Q$ in glass,so $v = +OQ$. Since $PO = OQ = x$,$v = +x$.The center of curvature is in the glass,so the radius of curvature $R$ is positive: $+R$.Substituting these values into the formula:$\frac{1.5}{x} - \frac{1.0}{-x} = \frac{1.5 - 1.0}{R}$$\frac{1.5}{x} + \frac{1.0}{x} = \frac{0.5}{R}$$\frac{2.5}{x} = \frac{0.5}{R}$$x = \frac{2.5}{0.5} 	imes R = 5R$.Thus,the distance $PO = 5R$.

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