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

Answer 1

The correct option is A $21.33cm$ above the lens

For convex lens,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{20}+\frac{1}{80}=\frac{1}{f}\Rightarrow f=16cm$
Now when water is poured, the image will shift. Its distance from the surface of water
$\frac{t}{m}=\frac{64\times 3}{4}=48cm$
Hence its distance from lens $=48+16=64cm$ . This will be the new object distance.
$\frac{1}{v}-\frac{1}{(-64)}=\frac{1}{16}$
Hence, $v=\frac{64}{3}=21.33cm$ above the lens.