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

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(D) $1$. First,calculate the apparent depth of the fish as seen from the air. The actual depth is $h = 0.4\, m$ and the refractive index of water is $

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The same location of the fish

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(D) $1$. First,calculate the apparent depth of the fish as seen from the air. The actual depth is $h = 0.4\, m$ and the refractive index of water is $\mu = 4/3$. The apparent depth $h'$ is given by $h' = h / \mu = 0.4 / (4/3) = 0.3\, m$.$2$. The distance of this apparent image from the lens is $u = -(0.2 + 0.3) = -0.5\, m$ (using sign convention where light travels from object to lens).$3$. Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $f = +3\, m$ and $u = -0.5\, m$:$\frac{1}{3} = \frac{1}{v} - \frac{1}{-0.5}$$\frac{1}{3} = \frac{1}{v} + 2$$\frac{1}{v} = \frac{1}{3} - 2 = -5/3$$v = -0.6\, m$.$4$. The negative sign indicates that the image is formed on the same side as the object,at a distance of $0.6\, m$ from the lens. Since the lens is $0.2\, m$ above the water,the image is $0.6 - 0.2 = 0.4\, m$ below the water surface,which is the same location as the fish.

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