WBJEE · Physics · Ray Optics
A thin convex lens is placed just above an empty vessel of depth \(80 \mathrm{~cm}\). The image of a coin kept at the bottom of the vessel is thus formed \(20 \mathrm{~cm}\) above the lens. If now water is poured in the vessel up to a height of \(64 \mathrm{~cm}\), what will be the approximate new position of the image. Assume that refractive index of water is \(4 / 3\).
- A \(21.33 \mathrm{~cm}\) above the lens
- B \(6.67 \mathrm{~cm}\) below the lens
- C \(33.67 \mathrm{~cm}\) above the lens
- D \(24 \mathrm{~cm}\) above the lens
Answer & Solution
Correct Answer
(A) \(21.33 \mathrm{~cm}\) above the lens
Step-by-step Solution
Detailed explanation
Hint: \(\mathrm{u}=-80 \mathrm{~cm} \quad \mathrm{v}=+20 \mathrm{~cm}\) \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{20}+\frac{1}{80}=\frac{5}{80}\) \(f=16 \mathrm{~cm}\) \(\mathrm{u}^{\prime}=16+\frac{64 \times 3}{4}=64 \mathrm{~cm}\)…
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