Question

Obtain an expression for the normal displacement of an object immersed in water.

Answer 1

When an object is placed in water and viewed from air, it appears to be raised due to refraction at the water-air interface. This vertical shift is called normal (apparent) displacement.

Let:

t = Real depth of the object (actual depth)

d = Apparent depth

n = Refractive index of water

From the formula for apparent depth:

d = `t/n`

Now, the normal displacement is the difference between real and apparent depth:

Δ = t − d

= `t - t/n`

= `t(1 - 1/n)`