A luminous point object O is placed at a dist | vedclass

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(A) Given: $n_1 = \frac{4}{3}$,$n_2 = \frac{3}{2}$,$R = 10 	ext{ cm}$.The object distance $u = -2R = -20 	ext{ cm}$ (using sign convention).The refr

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$30 	ext{ cm}$ in the rarer medium

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(A) Given: $n_1 = \frac{4}{3}$,$n_2 = \frac{3}{2}$,$R = 10 	ext{ cm}$.The object distance $u = -2R = -20 	ext{ cm}$ (using sign convention).The refraction formula at a spherical surface is given by:$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$Substituting the values:$\frac{3/2}{v} - \frac{4/3}{-20} = \frac{3/2 - 4/3}{10}$$\frac{3}{2v} + \frac{4}{60} = \frac{1/6}{10}$$\frac{3}{2v} + \frac{1}{15} = \frac{1}{60}$$\frac{3}{2v} = \frac{1}{60} - \frac{1}{15} = \frac{1 - 4}{60} = -\frac{3}{60} = -\frac{1}{20}$$\frac{3}{2v} = -\frac{1}{20}$$2v = -60 \implies v = -30 	ext{ cm}$.Since $v$ is negative,the image is formed on the same side as the object (in the rarer medium) at a distance of $30 	ext{ cm}$ from the pole $P$.

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