Q.An object of size $3.0\ \text{cm}$ is placed $14\ \text{cm}$ in front of a concave lens of focal length $21\ \text{cm}$. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

The key idea is refraction by a thin lens, using the lens formula and magnification to determine image nature, size, and position.

Step 1 — Apply the lens formula.

For a concave lens, $f = -21\ \text{cm}$ and $u = -14\ \text{cm}$ (sign convention: distances measured from lens, object side negative).

Step 2 — Find magnification and image size.

Image height $h_i = m \times h_o = 0.6 \times 3.0 = 1.8\ \text{cm}$.

Step 3 — Describe the image.

$v$ is negative → image is on the same side as the object (virtual). $m$ positive and less than 1 → image is erect and diminished.

Step 4 — Effect of moving object further away. …

Why the Lens Formula Works for Concave Lenses

A concave lens always diverges incoming light rays. No matter where you place the object in front of it, the refracted rays never actually meet — they only appear to come from a point on the same side as the object. That means the image is always virtual, erect, and diminished.

The standard lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ works for all thin lenses if you follow the sign convention strictly. For a concave lens, the focal length is taken as negative. The object distance $u$ is always negative (object is on the incident side). The image distance $v$ comes out negative, confirming the image is on the same side as the object — virtual.

Step-by-step solution

1. Write down the given data with proper signs

Object size $h_o = +3.0\ \text{cm}$ (positive by convention for erect object).

Object distance $u = -14\ \text{cm}$ (negative because object is in front of the lens, against incident light).

Focal length of concave lens $f = -21\ \text{cm}$ (negative for diverging lens).

2. Apply the lens formula to find $v$

Substitute:

That minus sign on $u$ is crucial — careful:

Now isolate $\frac{1}{v}$:

Find a common denominator (42):

So:

The negative sign tells us the image is on the same side as the object — virtual.

3. Find the magnification and image size

Magnification $m = \frac{h_i}{h_o} = \frac{v}{u}$

The positive sign means the image is erect (same orientation as object). The magnitude $0.6$ means the image is smaller than the object.

Image height:

4. Describe the image fully

The image is:

• Virtual (rays appear to diverge from it)
• Erect (same orientation as object)
• Diminished (size $1.8\ \text{cm}$)
• Located $8.4\ \text{cm}$ from the lens on the same side as the object

5. What happens when the object moves farther away?

Let the object distance increase (more negative, say $u = -30\ \text{cm}$). From the lens formula:

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} \quad \text{(with signs)}$ …

Method: Lens Formula & Magnification Approach

This is the standard method for solving image formation problems with thin lenses.

Step 1 – Identify given data and sign convention

We use the Cartesian sign convention (distances measured from optical centre, direction of incident light is positive):

• Object size: $h_o = +3.0\ \text{cm}$
• Object distance: $u = -14\ \text{cm}$ (negative because object is in front of lens)
• Focal length: $f = -21\ \text{cm}$ (negative for concave lens)

Step 2 – Apply the lens formula

The lens formula is:

Substitute values:

Take LCM (42):

Thus:

Image is formed $8.4\ \text{cm}$ in front of the lens (on the same side as the object).

Step 3 – Find magnification and image size

Magnification:

Image size:

Step 4 – Describe the image

• Position: $8.4\ \text{cm}$ in front of the lens (virtual image)
• Nature: Virtual (since $v$ is negative)
• Orientation: Erect (since $m$ is positive)
• Size: Diminished ($1.8\ \text{cm}$, smaller than object) …

Common Mistakes & How to Avoid Them

Mistake 1: Using the wrong sign convention for a concave lens

The error: Students often treat a concave lens like a convex lens and take $f = +21\ \text{cm}$. This leads to a completely wrong image position and nature.

Why it happens: The word "concave" is confused with "converging." A concave lens is diverging — its focal length is negative by the Cartesian sign convention.

How to avoid: Memorise this rule:

• Convex lens (converging): $f > 0$
• Concave lens (diverging): $f < 0$

For this problem:

$f = -21\ \text{cm}$, $u = -14\ \text{cm}$ (object on left, so negative).

Mistake 2: Forgetting that $u$ is always negative

The error: Plugging $u = +14\ \text{cm}$ into the lens formula.

Why it happens: The problem says "placed in front of" — students take this as positive. But in the Cartesian sign convention, distances measured against the incident light direction are negative.

How to avoid: Always draw a ray diagram with the incident light coming from left to right. The object is on the left of the lens, so $u$ is negative.

Mistake 3: Misapplying the lens formula

The error: Using $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ with wrong signs, or using the mirror formula by mistake.

How to avoid: The lens formula is:

For this problem:

The negative $v$ means the image is virtual and on the same side as the object.

Mistake 4: Getting magnification sign wrong

The error: Calculating $m = \frac{v}{u}$ and forgetting the sign, or using $m = -\frac{v}{u}$ (which is for mirrors).

How to avoid: For lenses, magnification is:

Here:

• Positive $m$ → image is erect (same orientation as object)
• $|m| < 1$ → image is diminished

Mistake 5: Not describing the image completely

The error: Only giving position or only giving size.

How to avoid: Always describe an image using four characteristics:

1. Position (where is it?)
2. Nature (real or virtual?)
3. Size (magnified, diminished, or same size?)
4. Orientation (erect or inverted?)

Answer for this problem: …