A spherical surface of radius of curvature 10 | vedclass

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(C) The formula for refraction at a spherical surface is $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Given: $\mu_1 = 3/2 = 1.5$,$\mu

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greater than $90\,cm$

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(C) The formula for refraction at a spherical surface is $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Given: $\mu_1 = 3/2 = 1.5$,$\mu_2 = 4/3$,$R = +10\,cm$ (since the centre lies in the denser medium $Y$,the surface is convex towards the rarer medium $X$).Let the object distance be $u = -x$ (where $x > 0$).Substituting the values: $\frac{4/3}{v} - \frac{1.5}{-x} = \frac{4/3 - 1.5}{10}$.$\frac{4}{3v} + \frac{1.5}{x} = \frac{1.333 - 1.5}{10} = \frac{-0.166}{10} = -1/60$.For a real image,$v > 0$. Thus,$\frac{4}{3v} = -\frac{1}{60} - \frac{1.5}{x}$.Since $v > 0$,the right side must be positive: $-\frac{1}{60} - \frac{1.5}{x} > 0$ is not possible with real $u$. Re-evaluating sign convention: If the surface is concave towards the object,$R = -10\,cm$.Then $\frac{4/3}{v} + \frac{1.5}{x} = \frac{4/3 - 1.5}{-10} = \frac{-1/6}{-10} = 1/60$.$\frac{4}{3v} = \frac{1}{60} - \frac{1.5}{x}$. For $v > 0$,$\frac{1}{60} > \frac{1.5}{x} \implies x > 1.5 	imes 60 \implies x > 90\,cm$.

$A$ spherical surface of radius of curvature $10\,cm$ separates two media $X$ and $Y$ of refractive indices $3/2$ and $4/3$ respectively. The centre of the spherical surface lies in the denser medium. An object is placed in medium $X$. For the image to be real,the object distance must be:

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