Q.A magician during a show makes a glass lens with $n = 1.47$ disappear in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water?

Concept: Index Matching — an object becomes invisible when its refractive index matches that of the surrounding medium, so no light is reflected or refracted at the interface.

Reasoning:

1. For the lens to "disappear," the liquid must have the same refractive index as the glass lens.
2. Given $n_{\text{lens}} = 1.47$, the liquid must also have $n_{\text{liquid}} = 1.47$. …

Why a lens “disappears”

You’ve seen this trick: a glass rod vanishes when placed in a beaker of oil. It’s not magic — it’s index matching. Light bends when it crosses a boundary between two materials with different refractive indices. If the indices are equal, light passes straight through with no bending, no reflection, and no refraction. The object becomes invisible because your eye sees no change in the path of light — it’s as if the object isn’t there.

For a lens to disappear completely in a liquid, the refractive index of the liquid must equal the refractive index of the lens material. That’s the core idea.

Step-by-step reasoning

1. What does “disappear” mean optically?

   When light travels from one medium to another, part of it reflects and part refracts. The amount of reflection depends on the difference in refractive indices. If the two indices are identical, there is no reflection and no refraction — the boundary becomes optically invisible.

2. Apply this to the lens and the liquid.

   The lens is made of glass with $n = 1.47$. For it to vanish, the surrounding liquid must have the same refractive index. So the liquid’s refractive index must be $1.47$.

3. Check if water could work.

   Water has a refractive index of about $1.33$ (for visible light). That’s significantly different from $1.47$. If you put a glass lens of $n = 1.47$ into water, the boundary would still bend light — the lens would remain visible, though perhaps slightly less distinct than in air. …

Method: Index Matching (Optical Invisibility)

Concept: When a transparent object is placed in a liquid with the same refractive index, light passes through without bending at the interface — the object becomes invisible.

Steps

1. Identify the condition for disappearance The lens disappears when there is no refraction at the lens-liquid boundary. This happens when:

2. Apply the given data The lens has $n = 1.47$. Therefore:

3. Compare with known liquids Water has refractive index $n_{\text{water}} \approx 1.33$ (for visible light). Since $1.47 \neq 1.33$, the liquid cannot be water. …

Common Mistakes on the “Disappearing Lens” Problem

This question tests Index Matching — the principle that when two materials have the same refractive index, light passes through their interface without bending or reflecting, making the object “invisible.”

Mistake 1: Assuming the liquid must be water because it’s a common liquid

Why it’s wrong:

Water has $n \approx 1.33$, but the lens has $n = 1.47$. These are not equal, so the lens would still be visible in water.

How to avoid:

• Always compare numerical values — don’t guess based on familiarity.
• Memorise common refractive indices:
  • Water: $1.33$
  • Glass (crown): $\approx 1.52$
  • Air: $1.00$
• Here, the liquid must have $n = 1.47$ exactly — water is not a match.

Mistake 2: Thinking the lens “disappears” because it dissolves or becomes transparent

Why it’s wrong:

The lens is solid glass — it does not dissolve. “Disappear” here means optically invisible due to index matching, not chemical change.

How to avoid:

• Understand the physics: no refraction or reflection at the interface when $n_\text{lens} = n_\text{liquid}$.
• The lens is still physically present, but light passes through as if it’s not there.

Mistake 3: Forgetting that the lens is made of glass with a specific $n$

Why it’s wrong:

Some students assume the lens has $n = 1.5$ (typical crown glass) and then say the liquid must have $n = 1.5$. But the problem gives $n = 1.47$ — use the given value, not a default.

How to avoid:

• Read the problem carefully — the refractive index of the lens is explicitly stated.
• Never substitute a “standard” value unless the problem says so.

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