A metal coin is at the bottom of a beaker filled to a height of 6 cm. The refractive index of the liquid is (4/3). To an observer looking above the surface of the liquid, the coin will appear raised up by

Question

A metal coin is at the bottom of a beaker filled to a height of 6 cm. The refractive index of the liquid is (4/3). To an observer looking above the surface of the liquid, the coin will appear raised up by (A) 4.5 cm (B) 6.75 cm (C) 1.5 cm (D) 7.5 cm

Tardigrade Admin · Accepted Answer

Here, μ=(4/3), Real depth =6 cm μ=( text Real depth / text Apparent depth ) Apparent depth =(6/(4 / 3))=(18/4)=4.5 cm The coin will appear to be raised =6 cm -4.5 cm =1.5 cm

Q. A metal coin is at the bottom of a beaker filled to a height of $6 c m$ . The refractive index of the liquid is (4/3). To an observer looking above the surface of the liquid, the coin will appear raised up by

4.5 cm

6.75 cm

1.5 cm

7.5 cm

Here, $μ = \frac{4}{3}$ , Real depth $= 6 c m$
$μ = \frac{Real depth}{Apparent depth}$
Apparent depth $= \frac{6}{( 4/3 )} = \frac{18}{4} = 4.5 c m$
The coin will appear to be raised $= 6 c m - 4.5 c m = 1.5 c m$

Q. A metal coin is at the bottom of a beaker filled to a height of 6cm. The refractive index of the liquid is (4/3). To an observer looking above the surface of the liquid, the coin will appear raised up by