Question

a coin is placedat the bottom of a beaker containing water(refractive index=4/3) to a depth 12 cm. By what height the coin appears to be raised when seen from vertically above?​

Answer 1

Answer:

the height the coin appears to be raised when seen from vertically above=3cm

Explanation:

given that,

reflective index =4/3

real depth=12cm

apparent depth=?

reflective index =RD/AD

4/3=12/AD

AD*4=12*3

AD=12*3/4

AD=3*3

AD=9cm

now,

we have to find the height of the coin appears to be rise when seen from vertically above.

=real depth - apparent depth

=12 - 9 =3cm

I hope it will benefit you buddy.

please make my answer brainliest..