A coin lies at the bottom of a tank containing water to a Dept of 130cm. If the refractive index of the water is 1.3, calculate the apparent displacement of the coin viewed vertically from above

Respuesta :

Answer: 100 cm

Explanation: When an object at the bottom of a tank with water is viewed vertically from above, to the observer, it seems the object is closer to the eye, i.e., the apparent depth of the object is less than the real depth the object is. This is caused by Refraction, which happens when an incident light passing through an interface between two different media causing the light to be deflected.

The relation between real and apparent depths is given by:

[tex]n=\frac{Dr}{Da}[/tex]

where

n is refractive index of the surface

Dr is real depth

Da is apparent depth

For the coin at the bottom of a tank, apparent depth is:

[tex]n=\frac{Dr}{Da}[/tex]

[tex]Da=\frac{Dr}{n}[/tex]

[tex]Da=\frac{130}{1.3}[/tex]

Da = 100

Apparent displacement of the coin viewed vertically from above is 100 cm.