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

Question

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

Tardigrade Admin · Accepted Answer

Here: Actual depth of liquid h=6 cm Refactive index of the liquid =(4/3) Using the relation μ=( text actual depth (h)/ text apparent depth (x))= or x=h × (3/4)=6 × (3/4)=4.5 cm Hence, the coin will appear at a depth of =6-4.5=1.5 cm

Q. A metal coin is at the bottom of a beaker filled with a liquid of refractive index $4/3$ to height of $6 c m$ . To an observer looking from above the surface of the liquid, coin will appear at a depth of :

7.5 cm

6.75 cm

4.5 cm

1.5 cm

Here : Actual depth of liquid $h = 6 c m$
Refactive index of the liquid $= \frac{4}{3}$
Using the relation
$μ = \frac{actual depth ( h )}{apparent depth ( x )} =$
or $x = h \times \frac{3}{4} = 6 \times \frac{3}{4} = 4.5 c m$
Hence, the coin will appear at a depth of
$= 6 - 4.5 = 1.5 c m$

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