Question

A thin convex lens is placed just above an empty vessel of depth $80cm$. The image of a coin kept at the bottom of the vessel is thus formed $20cm$ above the lens. If now water is poured in the vessel up to a height of $64cm$, what will be the approximate new position of the image. Assume that refractive index of water is $4/3$.

• A. 21.33 cm above the lens
• B. 6.67 cm below the lens
• C. 33.67 cm above the lens
• D. 24 cm above the lens

Answer 1

Answer: (A)

$u=-80cm$
$v=+20cm$
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{20}+\frac{1}{80}=\frac{5}{80}$
$f=16cm$
$u^{\prime }=16+\frac{64\times 3}{4}=64cm$
$\frac{1}{16}=\frac{1}{v}+\frac{1}{64}$
$\therefore \frac{1}{v}=\frac{1}{16}-\frac{1}{64}=\frac{3}{64}$
$v=21.33$ above the lens