A point object is placed at a distance of 15 | vedclass

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(A) Using the formula for refraction at a spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$Here,$\mu_1 = 1$ (for air),$

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$30 \, cm$ to the left

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(A) Using the formula for refraction at a spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$Here,$\mu_1 = 1$ (for air),$\mu_2 = 1.5$ (for glass),$u = -15 \, cm$,and $R = +30 \, cm$ (as the center of curvature is to the right of the pole).Substituting the values: $\frac{1.5}{v} - \frac{1}{-15} = \frac{1.5 - 1}{30}$$\frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{30}$$\frac{1.5}{v} = \frac{1}{60} - \frac{1}{15} = \frac{1 - 4}{60} = -\frac{3}{60} = -\frac{1}{20}$$v = 1.5 	imes (-20) = -30 \, cm$.The negative sign indicates that the image is formed $30 \, cm$ to the left of the pole.

$A$ point object is placed at a distance of $15 \, cm$ from a glass rod with a spherical end. The radius of curvature of the spherical end is $30 \, cm$. Find the position of the image formed.

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