As shown in the figure,region BCDEF and ABFG | vedclass

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(D) The object $O$ is at the midpoint of the region $BCEF$. The distance $DE = 20 \, cm$. Since $O$ is at the midpoint,the distance of $O$ from the cu

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$20 \, cm$ from point $D$

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(D) The object $O$ is at the midpoint of the region $BCEF$. The distance $DE = 20 \, cm$. Since $O$ is at the midpoint,the distance of $O$ from the curved surface $D$ is $u = -10 \, cm$.The refraction occurs at the curved surface $D$ with radius of curvature $R = -10 \, cm$ (as the center of curvature is to the left of $D$).Using the formula for refraction at a spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$Here,$\mu_1 = 2.0$ (medium where object is placed),$\mu_2 = 1.0$ (air),$u = -10 \, cm$,and $R = -10 \, cm$.Substituting the values:$\frac{1}{v} - \frac{2}{-10} = \frac{1 - 2}{-10}$$\frac{1}{v} + \frac{1}{5} = \frac{-1}{-10} = \frac{1}{10}$$\frac{1}{v} = \frac{1}{10} - \frac{1}{5} = \frac{1 - 2}{10} = -\frac{1}{10}$$v = -10 \, cm$.This means the image is formed $10 \, cm$ to the left of point $D$.

As shown in the figure,region $BCDEF$ and $ABFG$ have refractive indices $2.0$ and $1.5$ respectively. $A$ particle $O$ is kept at the midpoint of region $BCEF$. The final image of the object as seen by the eye is at a distance of:

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