Question

A glass slab of thickness 2.5 cm having refractive index 5/3 is kept on an ink spot. A transparent beaker of very thin bottom, containing water of refractive index 4/3 up to 8 cm, is kept on the glass block. Calculate apparent depth of ink spot when seen from the outside air

Answer 1

Answer:

The apparant depth of ink spot seen from outside air is 7.5 cm

Explanation:

If n is refractive index of transparent medium with respect to air

Then

$\boxed{n=\frac{\text{Real Depth}}{\text{Apparent Depth}}}$

Therefore, $\text{Apparent Depth}=\frac{\text{Real Depth}}{n}$

In the question the refractive indices of water and that of glass block are given

The resultant apparent depth will be the sum of the two depts

Thus apparent depth

$=\frac{\text{Real Depth of Glass Block}}{\text{Refractive index of glass block}} +\frac{\text{Real Depth of WaterBody}}{\text{Refractive index of Water Body}}$

$=\frac{2.5}{5/3} +\frac{8}{4/3} \text{ cm}$

$=(1.5+6) \text{ cm}$

$=7.5\text{ cm}$

Therefore, the apparant depth of ink spot seen from outside air is 7.5 cm

Answer 2

Answer:

Apparant depth = 7.5 cm.