Question

A mark is made on the bottom of a vessel and over this mark, a glass slab of thickness $3.5cm$ and refractive index $7/4$ is placed. Now water (refractive index, $4/3$ ) is poured into the vessel so that the surface of water is $8cm$ above the upper surface of the slab. Looking down normally through the water, the apparent depth of the mark below the surface of water will be

• A. 6.33 cm
• B. 7.5 cm
• C. 8 cm
• D. 10 cm

Answer 1

Answer: (C)

Apparent depth
$=\sum \left(\frac{t_i}{{\mu }_i}\right)=\frac{t_w}{{\mu }_w}+\frac{t_g}{{\mu }_g}$
$=\frac{8}{4/3}+\frac{3.5}{7/4}=\frac{24}{4}+\frac{14}{7}=6+2=8cm$.