1. Tardigrade
  2. Question
  3. Physics
  4. A mark is made on the bottom of a vessel and over this mark, a glass slab of thickness 3.5 cm 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 8 cm 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

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Q. A mark is made on the bottom of a vessel and over this mark, a glass slab of thickness $3.5\, cm$ 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 $8\, cm$ 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

Ray Optics and Optical Instruments

Solution:

Apparent depth
$=\sum\left(\frac{t_{i}}{\mu_{i}}\right)=\frac{t_{w}}{\mu_{w}}+\frac{t_{g}}{\mu_{g}}$
$=\frac{8}{4 / 3}+\frac{3.5}{7 / 4}=\frac{24}{4}+\frac{14}{7}=6+2=8 \,cm$.
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