Name: Snell's Window
Uploaded: 2019-07-19
Duration: 4 min 53 s
Description: Snell's window is a phenomenon by which an underwater viewer sees everything above the surface through a cone of light of width of about 96 degrees. Let's le...

Question

An object $O$ is kept in the air and a lens of focal length $10 cm (in air)$ is kept at the bottom of a container which is filled up to a height $44 cm$ by water. The refractive index of water is $\frac{4}{3}$ and that of glass is $\frac{3}{2}$ . The bottom of the container is closed by a thin glass slab of refractive index $\frac{3}{2}$ . Find the distance $(in cm)$ of the final image formed by the system from the bottom of the container (refer to figure shown).

Accepted Answer

Using lens maker formula, the refractive index of the lens is given as

$10 cm (in air)$ ,

where $\frac{3}{2}$ is the refractive index of the lens material with respect to medium, $R_{1}$ is the radius of the first surface of the lens and $R_{2}$ is the radius of the second surface of the lens.

For air,

$\frac{1}{f} = (\frac{1.5}{1} - 1) (\frac{1}{R} - \frac{1}{- R}) \Rightarrow \frac{1}{f} = (0.5) (\frac{2}{R}) \Rightarrow f = R = 10 cm$ .

For water,

$\frac{1}{f} = (\frac{1.5}{(\frac{4}{3})} - 1) (\frac{2}{R}) \Rightarrow \frac{1}{f} = (\frac{1}{8}) (\frac{2}{10}) \Rightarrow f = 40 cm$ .

The position of object will be the apparent position of object when viewed from water

Using, $\frac{n_{2}}{h_{a}} = \frac{n_{1}}{h_{act}}$ ,

where $n_{2}$ is the refractive index of medium from where the object is viewed and $n_{1}$ is the refractive index of the medium in which object is kept.

$\frac{(\frac{4}{3})}{h_{a}} = \frac{1}{12} \Rightarrow h_{a} = 16 cm$ .

The distance of the object from the lens is $u = - (44 + 16) cm$ and focal length of the lens in the water is $f = 40 cm$ .

Using, lens formula, we have,

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{- 60} + \frac{1}{40} \Rightarrow v = 120 cm$ .

If the medium below the lens had been water, then the final image formed was at $120 cm$ below the bottom container.

The apparent distance of image when rays pass from water to glass slab at the base will be given by

$\frac{n_{glass}}{h_{a}} = \frac{n_{water}}{h_{act}} \Rightarrow h_{a} = \frac{n_{glass}}{n_{water}} h_{act} = \frac{(\frac{4}{3})}{(\frac{3}{2})} 120 = 135 cm$ .

Finally, the rays will pass from the glass slab at the base to the air outside, then the apparent position of the image will be given by

$\frac{n_{air}}{h_{a}} = \frac{n_{glass}}{h_{act}} \Rightarrow h_{a} = \frac{n_{air}}{n_{glass}} h_{act} = \frac{(1)}{(\frac{3}{2})} 135 = 90 cm$ .

Hence, the final image will be formed at $90 cm$ below the bottom of the container.

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