Q.In many experimental set-ups the source and screen are fixed at a distance say $D$ and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

Concept: Microscope Magnification (Lens Displacement Method)

Let the fixed distance between object and screen be $D$, and the focal length of the lens be $f$ ($D > 4f$ for two real images).

1. For a thin lens, $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$, with $u + v = D$. Substituting $v = D - u$ gives:

This quadratic in $u$ has two real roots $u_1$ and $u_2$ (since discriminant $\Delta = D^2 - 4Df > 0$). Hence two distinct lens positions exist.

2. The two object distances satisfy $u_1 + u_2 = D$ and $u_1 u_2 = D f$. The distance between the two lens positions is: $|u_1 - u_2| = \sqrt{(u_1+u_2)^2 - 4u_1u_2} = \sqrt{D^2 - 4Df}$ …

This is the classic displacement method for finding the focal length of a lens. The key insight is that the lens formula is symmetric: if $u$ and $v$ are the object and image distances, swapping them gives the same $f$. When the total distance $D = u + v$ is fixed and larger than $4f$, the quadratic in $u$ has two real roots — these correspond to the two lens positions.

Let’s work through it step by step.

1. Set up the geometry. The object and screen are fixed, separated by distance $D$. The lens is placed between them. Let $u$ be the distance from the object to the lens, and $v$ the distance from the lens to the screen. Then

2. Apply the thin lens formula. For a real image on the screen, both $u$ and $v$ are positive. The lens formula is

Substitute $v = D - u$:

3. Obtain the quadratic in $u$. Combine the fractions:

Cross-multiply:

4. Two real positions exist when $D > 4f$. The discriminant is

For a real image to form, we need $\Delta > 0$, i.e. $D > 4f$. Then the two roots are

Notice that $u_1$ and $u_2$ are symmetric about $D/2$, and $u_1 + u_2 = D$, $u_1 u_2 = Df$.

5. Find the distance between the two lens positions. The lens positions differ by

This is the separation between the two points where the lens can be placed.

6. Find the ratio of image sizes. Magnification for a thin lens is $m = v/u$. For the first position: $m_1 = \frac{v_1}{u_1} = \frac{D - u_1}{u_1}.$ …

Method: The Displacement Method for Two Conjugate Lens Positions

Use this method whenever an object and a screen (or object and image plane) are held at a fixed total separation, and a single lens is slid between them to find where it forms a sharp image — a setup that generally has two valid lens positions.

Steps

Step 1: Set up one equation linking $u$ and $v$ to the fixed separation

If the object-to-screen distance is $D$ and the lens sits at object distance $u$ (image distance $v$), then

This is purely geometric and holds regardless of the lens formula.

Step 2: Substitute into the lens formula to get a single equation in one unknown

Use $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ together with $v = D - u$ to eliminate $v$, producing a quadratic in $u$ alone.

Step 3: Recognise why two solutions exist

A quadratic generically has two roots. Physically, these correspond to the lens being close to the object (small $u$, large $v$) or close to the screen (large $u$, small $v$) — both giving a real image at the same fixed screen position, because the lens equation is symmetric under swapping $u \leftrightarrow v$. Real, physically valid roots exist only when the discriminant is positive, which translates into a minimum-separation condition on $D$ in terms of $f$ (e.g. $D > 4f$ for a converging lens).

Step 4: Extract the two required quantities from the roots …