Q.A tank is filled with water to a height of $12.5\ \text{cm}$. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be $9.4\ \text{cm}$. What is the refractive index of water? If water is replaced by a liquid of refractive index $1.63$ up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Apparent depth relates to real depth by $n=h/h'$. For water: $n=12.5/9.4\approx1.33$.

With the liquid replaced (same height, $n=1.63$): new apparent depth $h'=12.5/1.63\approx7.67\ \text{cm}$.

The microscope must move from $9.4\ \text{cm}$ to $7.67\ \text{cm}$ — i.e. raised by $9.4-7.67=1.73\ \text{cm}$.

Setting up — apparent depth

For near-normal viewing, the refractive index of a medium relates real depth $h$ to apparent depth $h'$ by

Step 1 — refractive index of water

Real depth $h=12.5\ \text{cm}$, apparent depth (as measured by the microscope) $h'=9.4\ \text{cm}$.

This matches the well-known refractive index of water — a good sanity check.

Step 2 — apparent depth in the new liquid

The liquid is replaced, keeping the same height $h=12.5\ \text{cm}$, but now with $n_{liquid}=1.63$.

Step 3 — distance the microscope must move

The microscope was originally focused at $9.4\ \text{cm}$ below the surface. It must now be focused at $7.67\ \text{cm}$ below the surface — a smaller depth, because the denser liquid ($n=1.63>1.33$) raises the apparent position of the needle further.

Since the new apparent depth is smaller, the microscope must be moved upward (toward the surface) by this amount.

Method: Apparent Depth Formula (Index Matching)

This problem uses the apparent depth method for refractive index measurement. The key principle: when viewing an object through a transparent medium, the apparent depth is less than the real depth due to refraction.

Step-by-step solution

Step 1: Recall the formula

For a plane surface viewed normally (from directly above):

where $\mu$ is the refractive index of the medium.

Step 2: Find refractive index of water

Given:

• Real depth $= 12.5\ \text{cm}$
• Apparent depth $= 9.4\ \text{cm}$

Step 3: Find apparent depth for the new liquid

For the liquid with $\mu = 1.63$ and same real depth $= 12.5\ \text{cm}$:

Step 4: Calculate the distance the microscope must be moved

The microscope was initially focused at $9.4\ \text{cm}$ (apparent depth for water).

Now it must focus at $7.67\ \text{cm}$ (apparent depth for new liquid).

Since the new apparent depth is smaller, the microscope must be raised (moved upward) by:

Final Answer

• Refractive index of water: 1.33
• Microscope must be raised by 1.73 cm

Here are the common mistakes students make on this classic refractive index problem, and how to avoid each one.

1. Confusing Apparent Depth with Real Depth

The Mistake:

Students often swap the two values — using $12.5\ \text{cm}$ as apparent depth and $9.4\ \text{cm}$ as real depth.

Why it happens:

The problem states "apparent depth is measured to be $9.4\ \text{cm}$." Some students misread or assume the larger number must be the real depth.

How to avoid:

Always label clearly:

• Real depth ($h$) = actual physical height of water = $12.5\ \text{cm}$
• Apparent depth ($h'$) = what the microscope reads = $9.4\ \text{cm}$

Key formula:

$\mu = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{h}{h'}$

2. Forgetting That Refractive Index > 1 for Water

The Mistake:

A student computes $\frac{9.4}{12.5} = 0.752$ and writes that as the refractive index.

Why it happens:

They invert the fraction without checking if the answer makes physical sense.

How to avoid:

Remember: For a denser medium (like water), $\mu > 1$.

• If your answer is less than 1, you have swapped numerator and denominator.
• Always do a sanity check: water’s $\mu \approx 1.33$, so your answer should be close to that.

Correct calculation:

3. Mishandling the Second Part — Sign of the Shift

The Mistake:

Students compute the new apparent depth correctly but then give the wrong direction for the microscope movement (up vs. down).

Why it happens:

They forget that a higher refractive index makes the apparent depth smaller (the bottom looks shallower).

How to avoid:

• For $\mu = 1.63$, apparent depth $h'' = \frac{12.5}{1.63} \approx 7.67\ \text{cm}$
• Compare with the first apparent depth ($9.4\ \text{cm}$): $7.67 < 9.4$, so the needle appears higher (closer to the surface).
• Therefore, the microscope must be raised (moved upward) to refocus.

Distance moved:

4. Using the Wrong Formula for Shift

The Mistake:

Some students try to use the formula for lateral shift (for a glass slab) instead of apparent depth shift.

How to avoid:

For normal viewing (microscope looking straight down), the only shift is vertical:

But here, since you already have $h'$ for water, just compute the difference between the two apparent depths.

5. Rounding Too Early

The Mistake:

Rounding $12.5/9.4$ to $1.3$ in the first part, then using $1.3$ in the second part — leading to an inaccurate shift.

How to avoid:

Keep at least 3 significant figures throughout:

• $\mu = 1.3298 \approx 1.33$
• $h'' = 12.5 / 1.63 = 7.6687\ \text{cm}$
• Shift $= 9.4 - 7.6687 = 1.7313\ \text{cm} \approx \mathbf{1.73\ cm}$

Quick Summary Checklist

Final answers:

• $\mu_{\text{water}} \approx \mathbf{1.33}$
• Microscope must be moved upward by $\mathbf{1.73\ cm}$