Question

A pond of depth 20 cm is filled with water of refractive index 4/3. Calculate the apparent depth of tank when viewed normally?

Answer 1

here the depth(d) given is 20cm
and refractive index(R.I) is 4/3
apparent depth = d/(R.I) = 20/(4/3) = 20 (3/4) = 15cm

Answer 2

The apparent depth of a tank filled with a liquid of refractive index n, viewed normally, can be calculated using the formula:

d_apparent = d_real / n

Apparent depth =15 cm

where d_real is the real depth of the tank and n is the refractive index of the liquid.

In this case, the real depth of the tank is 20 cm and the refractive index of the water is 4/3. Therefore, the apparent depth of the tank can be calculated as:

d_apparent = 20 cm / (4/3) = 20 cm * (3/4) = 15 cm

So, the apparent depth of the tank when viewed normally is 15 cm. This is because light slows down in the water, causing the tank to appear shallower than it actually is. The amount by which it appears shallower is proportional to the refractive index of the liquid. The higher the refractive index, the more the light slows down and the more the tank will appear shallower.

To summarize, the apparent depth of the 20 cm deep tank filled with water of refractive index 4/3 is 15 cm when viewed normally. This is due to the slowing of light in the water, which changes the perceived depth of the tank.

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