When a convex lens is placed above an empty t | vedclass

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(D) Initially, with the empty tank, the object distance $u = -45 \ cm$ and the image distance $v = +36 \ cm$. Using the lens formula $\frac{1}{f} = \f

Vedclass · Accepted Answer

$1.366$

Vedclass · Answer

(D) Initially, with the empty tank, the object distance $u = -45 \ cm$ and the image distance $v = +36 \ cm$. Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{36} - \frac{1}{-45} = \frac{1}{36} + \frac{1}{45} = \frac{5+4}{180} = \frac{9}{180} = \frac{1}{20} \Rightarrow f = 20 \ cm$.When the liquid is poured to a depth of $40 \ cm$, the apparent depth of the mark is $d' = \frac{40}{\mu}$. The remaining distance from the lens to the liquid surface is $45 - 40 = 5 \ cm$. Thus, the new object distance is $u' = -(\frac{40}{\mu} + 5) \ cm$. The new image distance is $v' = +48 \ cm$.Using the lens formula again: $\frac{1}{f} = \frac{1}{v'} - \frac{1}{u'}$$\frac{1}{20} = \frac{1}{48} - \frac{1}{-(\frac{40}{\mu} + 5)} = \frac{1}{48} + \frac{1}{\frac{40}{\mu} + 5}$$\frac{1}{\frac{40}{\mu} + 5} = \frac{1}{20} - \frac{1}{48} = \frac{12 - 5}{240} = \frac{7}{240}$$\frac{40}{\mu} + 5 = \frac{240}{7} \approx 34.286$$\frac{40}{\mu} = 34.286 - 5 = 29.286$$\mu = \frac{40}{29.286} \approx 1.366$.

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