Question

Metal coin is at bottom of a beaker filled with refractive index =4/3 to height of 6 cm. To an observer looking from above surface of liquid coin will appear at a depth:

Answer 1

Answer:

Explanation:

Here : Actual depth of liquid h=6cm,

Refractive, index of the liquid =4/3

Using the relation

μ=actual depth(h)/apparent depth(x)

or x=h×3/4

     =6×3/4

     =4.5cm.

Answer 2

The observer looking from above the surface of the liquid will see the coin appear to be at a depth of 4 cm.

• This is because the refractive index of the liquid causes light to bend as it passes from the air into the liquid, making the coin appear to be closer to the surface of the liquid than it actually is.
• The amount of bending of light is proportional to the difference in refractive index between the air and the liquid.
• In this case, the refractive index of the liquid is 4/3, which is higher than the refractive index of air (which is 1), so the light will bend toward the normal.
• The formula for the apparent depth of an object in a refractive medium is given by: $d = d_o / n$

Where d is the apparent depth, d_o is the actual depth, and n is the refractive index of the medium.

Substituting the values, we get:

$d = 6 cm / 4/3 = 4 cm$

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