Q.A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam $12\ \text{cm}$ from P. At what point does the beam converge if the lens is

Concept: Refraction at a spherical surface / lens formula for a real object on the opposite side. Here the incident beam is converging toward P, so P acts as a virtual object for the lens (the rays would meet at P if the lens were absent).

Reasoning:

1. For a lens, the Cartesian sign convention: distances measured against the incident light are negative. The incident beam is coming from the left and converging to P on the right. The lens is placed $12\ \text{cm}$ before P, so the object distance $u = +12\ \text{cm}$ (positive because the object is virtual and lies on the right side).

2. Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.

   Substitute $u = +12\ \text{cm}$.

3. (a) Convex lens: $f = +20\ \text{cm}$.

   $\frac{1}{v} = \frac{1}{20} + \frac{1}{12} = \frac{3+5}{60} = \frac{8}{60} \Rightarrow v = 7.5\ \text{cm}$. …

Why a virtual object?

When a beam of light is converging toward a point P, it means the rays are heading to meet at P without any lens. If you place a lens before P (i.e., in the path of the converging beam), the rays haven't yet reached P — they are still converging. From the lens's perspective, the rays appear to be coming from a point beyond the lens, on the other side. That point is a virtual object: the rays are real and converging toward it, but they never actually get there because the lens intercepts them.

Step-by-step solution

1. Set up the sign convention

We use the Cartesian sign convention (the standard for most Indian boards):

• Distances measured from the optical centre of the lens.
• Distances in the direction of incident light are positive.
• Distances opposite to incident light are negative.
• Focal length of convex lens: $f = +20\ \text{cm}$.
• Focal length of concave lens: $f = -16\ \text{cm}$.

The incident beam is converging toward P, which is $12\ \text{cm}$ beyond the lens (on the other side). So the virtual object is at a distance $u = +12\ \text{cm}$ from the lens.

2. Lens formula

The lens formula is:

where $v$ is the image distance from the lens. A positive $v$ means the image forms on the opposite side (real image), negative $v$ means on the same side as the incident light (virtual image).

3. Part (a): Convex lens, $f = +20\ \text{cm}$

Substitute $u = +12\ \text{cm}$, $f = +20\ \text{cm}$:

$v = \frac{15}{2} = 7.5\ \text{cm}$ …

Method: Real Object / Virtual Object Approach for Lens Formula

This problem uses the lens formula with careful sign convention — the key insight is that when a converging beam meets a lens before reaching its focus, the point P acts as a virtual object for the lens.

Sign Convention (Cartesian)

• Distances measured from lens: positive in direction of incident light
• For a converging beam, the object lies beyond the lens → object distance $u$ is positive

Step-by-step solution

Step 1: Identify object nature

The beam converges at P without the lens. With the lens placed 12 cm before P, the rays are heading toward P behind the lens.

→ P is a virtual object for the lens.

→ Object distance:

Step 2: Apply lens formula

Step 3: Solve for each case

(a) Convex lens, $f = +20\ \text{cm}$

Interpretation: Positive $v$ means image forms on the same side as the incident light — i.e., 7.5 cm from lens toward P.

(b) Concave lens, $f = -16\ \text{cm}$

$\frac{1}{v} - \frac{1}{12} = \frac{1}{-16}$ …

Common Mistakes & How to Avoid Them

This is a classic "tricky" refraction problem because the light is converging before it hits the lens — it's not coming from a distant object. Here are the most common errors students make:

1. ✗ Assuming the object is at infinity

Mistake: Students see "convergent beam" and think $u = \infty$, then blindly use $1/f = 1/v$.

Why it's wrong: The beam is already converging to point P. The lens intercepts this converging beam before it reaches P. The point P acts like a virtual object for the lens.

✓ How to avoid:

• Draw the ray diagram. The rays are heading toward P but hit the lens first.
• The object for the lens is the point where the rays would have met — that's P.
• Since the rays are converging toward P before the lens, P is a real object? No — careful: the rays are incident on the lens while converging. This makes P a virtual object (rays are converging to a point behind the lens from the lens's perspective).

2. ✗ Wrong sign convention for $u$

Mistake: Using $u = +12\ \text{cm}$ (positive) because "object is on the left."

Why it's wrong: In the Cartesian sign convention (used in most Indian boards):

• Light travels from left to right.
• For a real object, $u$ is negative (object to the left of lens).
• Here, the rays are converging toward P which is on the right of the lens. The object (P) is on the right side of the lens — so $u$ is positive.

✓ How to avoid:

• Always ask: "Where is the object relative to the lens?"
• If the object is on the right (opposite to incident light direction), $u$ is positive.
• For this problem: $u = +12\ \text{cm}$.

3. ✗ Using the lens formula incorrectly for virtual objects

Mistake: Plugging $u = +12$ into $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ without checking sign of $f$.

Why it's wrong: The lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ (with sign convention). For a convex lens, $f$ is positive; for concave, $f$ is negative.

✓ How to avoid:

• Memorise:
  • Convex lens: $f > 0$
  • Concave lens: $f < 0$
• Then solve:
  • (a) Convex: $\frac{1}{20} = \frac{1}{v} - \frac{1}{+12}$ → $\frac{1}{v} = \frac{1}{20} + \frac{1}{12}$ → $v = +7.5\ \text{cm}$
  • (b) Concave: $\frac{1}{-16} = \frac{1}{v} - \frac{1}{+12}$ → $\frac{1}{v} = -\frac{1}{16} + \frac{1}{12}$ → $v = +48\ \text{cm}$

4. ✗ Misinterpreting the sign of $v$

Mistake: Getting $v = +7.5\ \text{cm}$ and saying "image is 7.5 cm to the left of lens."

Why it's wrong: A positive $v$ means the image is on the right side of the lens (the same side as the virtual object).

✓ How to avoid:

• Sign convention:
  • $v > 0$ → image on the right (real image for real object, but here it's a real image formed on the opposite side)
  • $v < 0$ → image on the left (virtual image) …