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(D) Let $H_{R1}$ be the real depth of the liquid in the first case and $H_{app1}$ be the apparent depth. The microscope reading for the mark on the ta

Vedclass · Accepted Answer

$\frac{d - b}{d - c - b + a}$

Vedclass · Answer

(D) Let $H_{R1}$ be the real depth of the liquid in the first case and $H_{app1}$ be the apparent depth. The microscope reading for the mark on the table is $a$ and for the surface is $b$. Thus,the apparent depth $H_{app1} = b - a$.We know that refractive index $\mu = \frac{	ext{Real Depth}}{	ext{Apparent Depth}}$,so $H_{R1} = \mu(b - a)$  .......$(1)$In the second case,after adding more liquid,the reading for the mark is $c$ and for the surface is $d$. The apparent depth $H_{app2} = d - c$.Thus,$H_{R2} = \mu(d - c)$  .......$(2)$The difference in real depths is the additional liquid added,which corresponds to the change in the surface level reading: $H_{R2} - H_{R1} = d - b$  .......$(3)$Substituting $(1)$ and $(2)$ into $(3)$:$\mu(d - c) - \mu(b - a) = d - b$$\mu(d - c - b + a) = d - b$$\mu = \frac{d - b}{d - c - b + a}$

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