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(B) The apparent depth of the ink mark due to the glass slab is given by $d' = \frac{t}{\mu}$, where $t = 3 \ cm$ is the thickness of the slab and $\m

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(B) The apparent depth of the ink mark due to the glass slab is given by $d' = \frac{t}{\mu}$, where $t = 3 \ cm$ is the thickness of the slab and $\mu$ is its refractive index.The observer is at a distance of $5 \ cm$ above the slab.Therefore, the total apparent distance of the mark from the observer is the sum of the distance above the slab and the apparent depth of the mark inside the slab.Total apparent distance = $5 \ cm + \frac{3 \ cm}{\mu} = 4 \ cm$ is incorrect based on the problem statement. Let's re-evaluate: The mark is at the bottom of the slab. The apparent depth of the mark as seen from the top surface is $d_{app} = \frac{t}{\mu} = \frac{3}{\mu}$.The observer is $5 \ cm$ above the top surface.So, the total apparent distance from the observer is $5 + \frac{3}{\mu} = 4$ is not possible as the mark must appear closer. The problem implies the total distance from the observer to the apparent position is $4 \ cm$.If the observer is $5 \ cm$ above the slab, and the mark appears at a distance of $4 \ cm$ from the observer, then the apparent depth $d_{app}$ from the top surface must be $4 - 5 = -1 \ cm$, which is physically impossible. Re-reading: "For a person looking at the mark at a distance $5.0 \ cm$ above it". This likely means the total distance from the observer to the mark is $5 \ cm$ (including the slab). Let the total real distance be $5 \ cm$. The slab thickness is $3 \ cm$. The distance above the slab is $5 - 3 = 2 \ cm$.The apparent depth of the mark from the top surface is $\frac{3}{\mu}$.The total apparent distance from the observer is $2 + \frac{3}{\mu} = 4 \ cm$.$\frac{3}{\mu} = 4 - 2 = 2 \ cm$.$\mu = \frac{3}{2} = 1.5$.

$A$ glass slab of thickness $3 \ cm$ is placed on an ink mark on a piece of paper. For a person looking at the mark from a distance of $5.0 \ cm$ above the top surface of the slab, the mark appears to be at a distance of $4.0 \ cm$ from the observer. The refractive index of the slab is:

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