Q.A small pin fixed on a table top is viewed from above from a distance of $50\ \text{cm}$. By what distance would the pin appear to be raised if it is viewed from the same point through a $15\ \text{cm}$ thick glass slab held parallel to the table? Refractive index of glass $= 1.5$. Does the answer depend on the location of the slab?

Concept: Apparent shift due to refraction through a parallel slab — the image is formed closer to the observer by an amount equal to the real thickness reduced by the factor of refractive index.

Reasoning:

1. For a plane parallel slab, the apparent shift (the distance the object seems to move toward the observer) is given by:

where $t$ is the thickness of the slab and $\mu$ is the refractive index.

2. Here, $t = 15\ \text{cm}$, $\mu = 1.5$. Substituting: $\text{Shift} = 15\left(1 - \frac{1}{1.5}\right) = 15\left(1 - \frac{2}{3}\right) = 15 \times \frac{1}{3} = 5\ \text{cm}$ …

When you look at an object through a transparent slab, the light bends twice — once entering the slab and once leaving it. For a parallel-sided slab, these two refractions cancel out the angular deviation, but they do cause a lateral shift in the apparent position of the object. The net effect is that the object appears closer to you by a fixed amount.

This shift depends only on the thickness of the slab and its refractive index, not on where you hold the slab between you and the object. That’s a key point — and often a surprise.

Let’s work it out.

1. The formula for normal shift When you view an object through a parallel slab normally (i.e., along the perpendicular), the apparent shift towards the observer is:

where $t$ is the thickness of the slab and $\mu$ is its refractive index relative to the surrounding medium (air, $\mu_{\text{air}} \approx 1$).

Why this formula? Imagine the object at the bottom of the slab. Light from it travels through the slab, then through air to your eye. Because glass is optically denser, the rays bend away from the normal on exiting, making the object appear shallower. The derivation uses Snell’s law for small angles (paraxial approximation) and geometry — but the result is clean and exam-relevant.

2. Plug in the numbers Given: $t = 15\ \text{cm}$, $\mu = 1.5$.

So the pin appears raised by 5 cm.

3. Does the slab’s location matter? …

Method: Apparent Shift Due to a Parallel Slab (Normal Shift Formula)

This problem uses the normal shift concept — when an object is viewed through a parallel-sided slab, the image appears closer to the observer by a fixed amount.

Steps

1. Identify the given data

   • Thickness of glass slab, $t = 15\ \text{cm}$
   • Refractive index of glass, $\mu = 1.5$
   • Viewing is normal (from above, perpendicular to slab)

2. Recall the formula for apparent shift

   For a slab of thickness $t$ and refractive index $\mu$, when viewed normally:

This is the distance by which the object appears to be raised toward the observer.

3. Substitute the values

4. Answer the second part …

Common Mistakes & How to Avoid Them

Mistake 1: Using the Spherical Surface Formula Instead of the Slab Formula

The error: Students see "refraction" and immediately apply:

But a parallel slab has flat surfaces ($R = \infty$), so this formula is not applicable here.

How to avoid:

• Identify the shape: Is the surface curved or flat?
• For a parallel slab, use the apparent shift formula directly:

where $t$ = thickness, $n$ = refractive index.

Mistake 2: Forgetting That Shift Depends Only on Thickness and $n$

The error: Students try to use distances like 50 cm in the shift calculation.

The truth: The apparent shift for a slab viewed normally is independent of the viewing distance (as long as the slab is between the eye and the object).

How to avoid:

• Memorise: For normal viewing through a slab, shift = $t(1 - 1/n)$.
• The 50 cm distance is irrelevant to the shift magnitude — it only tells you the viewing is from above (normal incidence).

Mistake 3: Incorrect Sign Convention for Shift Direction

The error: Some students add the shift instead of subtracting it.

The correct logic:

• The pin is raised (appears closer to the observer).
• Shift = $t - t_{\text{apparent}} = t - \frac{t}{n} = t\left(1 - \frac{1}{n}\right)$

How to avoid:

• Draw a ray diagram: light bends away from the normal when entering a rarer medium → image shifts upward (toward the observer).
• Always check: Does the answer make physical sense? (Yes, the pin appears raised.)

Mistake 4: Answering the "Does it depend on slab location?" Part Incorrectly

The error: Students say "Yes, because distance from eye changes."

The correct answer:

No, the apparent shift does not depend on where the slab is placed between the eye and the object — as long as the slab is parallel and viewed normally.

Why:

• The shift is purely due to the slab's thickness and refractive index. …