An object is placed at a distance of 20 , cm | vedclass

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(B) The formula for refraction at a spherical surface is given by: $\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R}$Given:$n_{1} = 1$ (ref

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$40 \, cm$ from the pole and inside the denser medium

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(B) The formula for refraction at a spherical surface is given by: $\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R}$Given:$n_{1} = 1$ (refractive index of rarer medium)$n_{2} = 2$ (refractive index of denser medium)$u = -20 \, cm$ (object distance,following sign convention)$R = +10 \, cm$ (radius of curvature for a convex surface)Substituting the values into the formula:$\frac{2}{v} - \frac{1}{-20} = \frac{2 - 1}{10}$$\frac{2}{v} + \frac{1}{20} = \frac{1}{10}$$\frac{2}{v} = \frac{1}{10} - \frac{1}{20}$$\frac{2}{v} = \frac{2 - 1}{20} = \frac{1}{20}$$v = 40 \, cm$Since $v$ is positive,the image is formed at a distance of $40 \, cm$ from the pole inside the denser medium.

An object is placed at a distance of $20 \, cm$ in a rarer medium from the pole of a convex spherical refracting surface of radius of curvature $10 \, cm$. If the refractive index of the rarer medium is $1$ and that of the denser medium is $2$,then the position of the image is at

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