A small fish,0.4,m below the surface of a lak | vedclass

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(B) Step $1$: Find the apparent depth of the fish due to refraction at the water surface.Using the formula for refraction at a plane surface: $\frac{\

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$A$ distance of $0.6\,m$ from the water surface

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(B) Step $1$: Find the apparent depth of the fish due to refraction at the water surface.Using the formula for refraction at a plane surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = 0$,where $\mu_1 = \frac{4}{3}$,$\mu_2 = 1$,and $u = -0.4\,m$.$\frac{1}{v} - \frac{4/3}{-0.4} = 0 \Rightarrow \frac{1}{v} = -\frac{4/3}{0.4} = -\frac{4/3}{2/5} = -\frac{10}{3}$.So,$v = -0.3\,m$. The apparent depth is $0.3\,m$ below the water surface.Step $2$: Use the lens formula to find the final image position.The distance of the apparent image from the lens is $u' = -(0.3 + 0.2) = -0.5\,m$.The focal length of the converging lens is $f = +3\,m$.Using the lens formula: $\frac{1}{v'} - \frac{1}{u'} = \frac{1}{f}$.$\frac{1}{v'} - \frac{1}{-0.5} = \frac{1}{3} \Rightarrow \frac{1}{v'} + 2 = \frac{1}{3}$.$\frac{1}{v'} = \frac{1}{3} - 2 = -\frac{5}{3}$.$v' = -0.6\,m$.The negative sign indicates that the image is virtual and formed $0.6\,m$ above the lens. Since the lens is $0.2\,m$ above the water surface,the image is at a distance of $0.6\,m$ above the lens,which is $0.6\,m$ from the lens position.

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