Q.A screen is placed $90\ \text{cm}$ from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by $20\ \text{cm}$. Determine the focal length of the lens.

Concept: Displacement method for focal length of a convex lens.

When object and screen are fixed at a distance $D$ apart, a convex lens forms a sharp image at two positions separated by $d$. The focal length is given by:

Step 1: Identify $D$ and $d$.

$D = 90\ \text{cm}$ (distance between object and screen).

$d = 20\ \text{cm}$ (separation between the two lens positions).

Step 2: Substitute into the formula. …

The displacement method for finding the focal length of a convex lens is a classic experiment — and a favourite in exams — because it avoids the need to measure object and image distances separately. The key insight: when the object and screen are fixed at a separation $D$ greater than $4f$, there are two distinct lens positions that produce a sharp image on the screen. One gives a magnified image, the other a diminished one. The distance between these two positions, $d$, together with $D$, directly gives $f$.

Why does this happen? For a given object distance $u$, the lens formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ with $v = D - u$ becomes a quadratic in $u$. Two real roots exist when $D > 4f$, and the difference between them is exactly $d = \sqrt{D^2 - 4fD}$. Rearranging gives the neat formula.

Let’s work through it step by step.

1. Set up the geometry. Object and screen are fixed $90\ \text{cm}$ apart. So $D = 90\ \text{cm}$. Let the lens be at a distance $u$ from the object. Then the image distance from the lens is $v = D - u$ (since the screen is on the other side). The lens formula:

2. Form the quadratic in $u$. Combine the fractions:

So

Multiply by $-1$:

3. Two solutions — the two lens positions. This quadratic has two roots $u_1$ and $u_2$ (the two object distances for which a sharp image forms). Their sum and product:

The distance between the two lens positions is $d = |u_1 - u_2|$.

Using the identity $(u_1 - u_2)^2 = (u_1 + u_2)^2 - 4u_1 u_2$, we get

4. Solve for $f$. Rearranging: …

Method: Displacement Method (for Convex Lens)

This method is used when a convex lens forms a real image of an object on a fixed screen at two different positions — a classic exam setup.

Step 1: Understand the given data

• Distance between object and screen: $D = 90\ \text{cm}$
• Distance between the two lens positions: $d = 20\ \text{cm}$

Step 2: Recall the formula

For the displacement method, the focal length $f$ of the convex lens is given by:

This formula comes from the fact that for a fixed object-screen distance $D$, the lens forms a sharp image at two positions separated by $d$, and the lens formula is applied to both positions.

Step 3: Substitute the values

Step 4: Simplify

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Here are the common mistakes students make on this classic displacement method problem, along with how to avoid each.

1. Confusing the two lens positions

Mistake:

Students think the two positions are symmetric about the midpoint, or they try to guess which distance is $u$ and which is $v$ without using the formula.

Why it happens:

The problem says the lens forms a sharp image at two positions separated by $20\ \text{cm}$. Many assume one position is obvious — it is not.

How to avoid:

Use the displacement method formula directly:

If $D$ = distance between object and screen, and $d$ = separation between the two lens positions, then

$f = \frac{D^2 - d^2}{4D}$

Here:

• $D = 90\ \text{cm}$
• $d = 20\ \text{cm}$

So:

Key takeaway: Memorise the formula — it saves time and avoids confusion.

2. Forgetting that $u$ and $v$ swap between positions

Mistake:

Students set up $u$ and $v$ for one position, then use the same values for the second position.

Why it happens:

They don’t realise that in the displacement method, the two positions correspond to interchanging object and image distances.

How to avoid:

Remember:

• At position 1: $u_1$, $v_1$
• At position 2: $u_2 = v_1$, $v_2 = u_1$

This is the core symmetry. The lens formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ gives the same $f$ for both.

3. Using the lens formula incorrectly with the given numbers

Mistake:

Plugging $u = 90$ or $v = 90$ directly into $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.

Why it happens:

They think the 90 cm is either $u$ or $v$, but it’s actually $u + v = D$.

How to avoid:

Always write:

Then use the fact that the two positions give $u_1 - u_2 = d$ (or $v_1 - v_2 = d$).

Solve the system:

This gives $v = 55$, $u = 35$ (or vice versa). Then:

4. Sign convention errors

Mistake: …