An object is placed at the focus of concave lens having focal length f. What is the magnification an

Q: Why is the focal length of a concave lens taken as negative?

By sign convention, the principal focus of a concave lens lies on the same side as the incident light (left side), making its coordinate negative.

Q: Why is the object distance $u$ negative?

According to the Cartesian sign convention, all distances measured in the direction opposite to the incident light (typically to the left of the optical centre) are taken as negative.

Q: What does a positive magnification mean?

A positive magnification indicates that the image formed is virtual and erect, which is always true for a real object placed in front of a concave lens.

Q: Can a concave lens ever form a real image?

For a real object, a concave lens always forms a virtual, erect, and diminished image. It can only form a real image if the object is virtual (converging rays incident on the lens).

Q: Why did we take the absolute value of $v$ for the final answer?

The question asks for the 'distance' of the image from the optical centre. Distance is a scalar quantity and is always positive, so we take the magnitude $|v| = f/2$.

Question

An object is placed at the focus of concave lens having focal length $f$. What is the magnification and distance of the image from the optical centre of the lens?

Options:

(A) $1, \infty$

(B) Very high, $\infty$

(C) $\frac{1}{2}, \frac{f}{2}$

(D) $\frac{1}{4}, \frac{f}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in an optics lab, holding a **concave lens**. You place an object exactly at its principal focus. The question asks us to find two things: the exact distance where the image will form, and how much it will be magnified.

A concave lens is a diverging lens. It takes parallel rays of light and spreads them out. Because of this, it has a reputation for always forming **virtual, erect, and diminished** images for any real object. But let's prove this mathematically.

### The Master Equation

To unlock the secrets of this optical system, we need our master tool: the **Lens Formula**. It beautifully connects the object distance $u$, the image distance $v$, and the focal length $f$ through the relation:
$$ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} $$

We also need the formula for linear magnification $m$, which tells us how large the image is compared to the object. For lenses, it is simply the ratio of the image distance to the object distance:
$$ m = \frac{v}{u} $$

### Applying Sign Conventions

Before we plug in the numbers, we must strictly follow the **Cartesian sign convention**. This is where many students make a silly mistake!

For a concave lens, the principal focus lies on the left side (the side from which light originates). Therefore, its focal length is taken as negative, so we use $-f$.

The object is also placed on the left side, exactly at the focus. This means the object distance $u$ is also negative, giving us $u = -f$.

### The Final Calculation

Now, let's substitute these values into our lens formula. 
$$ \frac{1}{v} - \frac{1}{-f} = \frac{1}{-f} $$

Moving the terms around, we get:
$$ \frac{1}{v} = -\frac{1}{f} - \frac{1}{f} $$
$$ \frac{1}{v} = -\frac{2}{f} $$

Flipping this over, we find the image distance:
$$ v = -\frac{f}{2} $$

The negative sign tells us that the image forms on the **left side** of the lens, exactly halfway between the optical centre and the focus. Since distance is a positive quantity, the distance is $\frac{f}{2}$.

Finally, let's calculate the magnification:
$$ m = \frac{v}{u} = \frac{-f/2}{-f} = \frac{1}{2} $$

The magnification is $+\frac{1}{2}$. The positive sign confirms the image is **erect and virtual**, and the value $\frac{1}{2}$ tells us it is exactly **half the size** of the original object. The physics perfectly matches the math!

Animated Solution for Physics - Optics: An object is placed at the focus of concave lens having focal length $f$ . What is the magnification and distance of the image from the optical centre of the lens?

Select Answer:

Visualized Solution

Visual Anchor

Logic Bridge

Raw Setup

Atomic Compute: Image Distance

Atomic Compute: Magnification

Final Answer

The Way Forward

The Sigma Insight: Lens

Analyzing the Setup

The Master Equation

Applying Sign Conventions

The Final Calculation

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