A fish is at the centre of a spherical water- | vedclass

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(B) We use the refraction formula at a spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_{surf}}$.Here,the light travels

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$2$

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(B) We use the refraction formula at a spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_{surf}}$.Here,the light travels from the child (in air,$\mu_1 = 1$) to the fish (in water,$\mu_2 = 4/3$).The object (child's nose) is at a distance $u = -(2R - R) = -R$ from the pole $P$ of the spherical surface.The radius of curvature of the surface is $R_{surf} = -R$ (as measured against the direction of incident light).Substituting the values:$\frac{4/3}{v} - \frac{1}{-R} = \frac{4/3 - 1}{-R}$$\frac{4}{3v} + \frac{1}{R} = \frac{1/3}{-R}$$\frac{4}{3v} = -\frac{1}{3R} - \frac{1}{R} = -\frac{4}{3R}$$\frac{1}{v} = -\frac{1}{R} \Rightarrow v = -R$.This image is formed at a distance $R$ from the pole $P$ inside the bowl. Since the fish is at the centre (distance $R$ from $P$),the total distance of the image from the centre is $R + R = 2R$.

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