Q.The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall $3\ \text{m}$ away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

The key idea is that for a real image to be formed on a screen, the object and image distances must be positive. The lens formula gives a constraint on the focal length when the total distance between object and image is fixed.

Reasoning:

1. Let the distance between the object (bulb) and the image (opposite wall) be $D = 3\ \text{m}$. If the lens is placed at a distance $u$ from the object, then the image distance is $v = D - u$.

2. Using the lens formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$, we get:

   $\frac{1}{f} = \frac{1}{u} + \frac{1}{D-u} = \frac{D}{u(D-u)}.$ …

The problem is about forming a real image of a real object using a convex lens. The object (the bulb) is fixed on one wall, and the image must be formed on the opposite wall, which is 3 m away. So the total distance between object and image is fixed at 3 m. The lens can be placed anywhere between them.

The key idea: For a convex lens, a real image is formed on the opposite side of the lens from the object. The lens formula is

with the sign convention where $u$ is negative (object distance measured from lens, opposite to incident light direction) and $v$ is positive (real image on the other side). But it's simpler here to use magnitudes: let $u$ be the distance from lens to object (positive), and $v$ be the distance from lens to image (positive). Then the lens formula becomes

and the total distance $u + v = 3\ \text{m}$.

We want the maximum possible focal length $f$ for which a real image can be formed. Let's work through it.

1. Set up the relation. Given $u + v = 3$, we have $v = 3 - u$. Substitute into the lens formula:

This is valid only when $u > 0$ and $v > 0$, i.e., $0 < u < 3$.

2. Express $f$ as a function of $u$. Combine the fractions:

So

This is a quadratic in $u$: $f = \frac{1}{3}(3u - u^2)$.

3. Find the maximum of $f$. The expression $u(3 - u)$ is a downward-opening parabola in $u$, with maximum at the vertex. For a quadratic $au^2 + bu + c$, the vertex is at $u = -\frac{b}{2a}$. Here $u(3 - u) = -u^2 + 3u$, so $a = -1$, $b = 3$, giving …

Method: Lens Formula with Real Object and Real Image Condition

This problem uses the Lens Formula approach, combined with the constraint that both object and image are real (on opposite walls).

Step-by-step solution

Step 1: Identify the given data

• Distance between object and image: $D = 3\ \text{m}$
• Both object and image are real (on opposite walls)
• Lens is convex (converging)

Step 2: Apply the lens formula

The lens formula is:

Where:

• $u$ = object distance (negative by sign convention for real object)
• $v$ = image distance (positive for real image)

Step 3: Express the distance constraint

If the lens is placed between the walls, and total distance is $D$:

Let $u = -x$ (where $x > 0$ is the distance from lens to object wall). Then:

Step 4: Substitute into lens formula

Step 5: Find condition for maximum focal length

For a real image to form, we need $f$ to be real and positive. The focal length is maximum when the denominator is minimum.

Rewrite:

So:

$f = \frac{x(D - x)}{D}$ …

Here are the common mistakes students make on this classic refraction/lens problem, and how to avoid each.

Mistake 1: Forgetting the Lens Formula Sign Convention

The error:

Students plug $u = -3\ \text{m}$ and $v = +3\ \text{m}$ directly into the lens formula without realising that the object and image are on opposite sides of the lens. They often treat both distances as positive, leading to a wrong focal length.

Why it happens:

They confuse the real object / real image situation with a simple "distance between object and image" approach.

How to avoid:

Always use the Cartesian sign convention:

• Object distance $u$ is negative (real object, left of lens).
• Image distance $v$ is positive (real image, right of lens).

So here:

• $u = -x$ (distance from lens to object)
• $v = +(3 - x)$ (distance from lens to image, since total wall-to-wall distance is $3\ \text{m}$)

Mistake 2: Assuming Object and Image are at Equal Distances

The error:

Many students assume $|u| = |v| = 1.5\ \text{m}$ because the total distance is $3\ \text{m}$. They then compute $f = 0.75\ \text{m}$ and stop.

Why it happens:

They think "symmetry" always gives the maximum focal length — but this is only true if the lens can be placed anywhere.

How to avoid:

Realise that the maximum focal length occurs when the lens is placed exactly midway only if you want a specific magnification. For maximum $f$, you must vary the lens position and find the condition where $f$ is largest.

Mistake 3: Not Using the Condition for Real Image Formation

The error:

Students forget that for a real image, the lens formula must yield a positive $f$, and the quadratic in $x$ must have real roots.

Why it happens:

They treat the problem as a simple substitution rather than an optimisation.

How to avoid:

Write the lens formula:

Substitute $u = -x$, $v = 3 - x$:

Then find the minimum of $\frac{1}{f}$ (which gives maximum $f$) by differentiating or using AM–GM inequality.

Mistake 4: Algebraic Errors in the Quadratic

The error:

When simplifying $\frac{1}{f} = \frac{3}{x(3-x)}$, students often write $f = \frac{x(3-x)}{3}$ but then forget that $f$ is maximum when the denominator $x(3-x)$ is maximum. …