A thin convex lens is placed just above an em | vedclass

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Question

(A) Step $1$: Find the focal length of the lens using the empty vessel condition.For an empty vessel, the object distance $u = -80 \,cm$ and the image

Vedclass · Accepted Answer

$21.33 \,cm$ above the lens

Vedclass · Answer

(A) Step $1$: Find the focal length of the lens using the empty vessel condition.For an empty vessel, the object distance $u = -80 \,cm$ and the image distance $v = +20 \,cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{20} - \frac{1}{-80} = \frac{1}{20} + \frac{1}{80} = \frac{4+1}{80} = \frac{5}{80} = \frac{1}{16} \,cm^{-1}$.So, the focal length $f = 16 \,cm$.Step $2$: Find the new object distance after pouring water.When water of height $h = 64 \,cm$ is poured, the apparent depth of the coin is $d' = \frac{h}{\mu} = \frac{64}{4/3} = 64 	imes \frac{3}{4} = 48 \,cm$.The distance of the coin from the lens is $80 \,cm$. The distance of the apparent image from the bottom is $64 - 48 = 16 \,cm$. Thus, the new object distance $u'$ from the lens is $80 - 16 = 64 \,cm$ (or $16 \,cm$ air + $48 \,cm$ apparent depth).So, $u' = -64 \,cm$.Step $3$: Find the new image position $v'$.Using the lens formula $\frac{1}{f} = \frac{1}{v'} - \frac{1}{u'}$:$\frac{1}{16} = \frac{1}{v'} - \frac{1}{-64} \implies \frac{1}{v'} = \frac{1}{16} - \frac{1}{64} = \frac{4-1}{64} = \frac{3}{64}$.$v' = \frac{64}{3} \approx 21.33 \,cm$.Therefore, the image is formed $21.33 \,cm$ above the lens.

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