Worked Example: Image Distance of a Thin Lens

This introductory example demonstrates the solution format using the Gaussian thin-lens equation. It is self-contained and is not copied from a textbook.

Problem in our own words

An object is \(300\ \mathrm{mm}\) in front of a converging thin lens whose focal length is \(100\ \mathrm{mm}\). Find the image distance and state whether the image is real or virtual.

What is known

We use the real-is-positive convention for this example. A converging lens has positive focal length, and a real object placed in front of the lens has positive object distance:

\[f = +100\ \mathrm{mm}, \qquad d_o = +300\ \mathrm{mm}, \qquad d_i = \;?\]

All distances already use millimetres, so no unit conversion is needed.

Step 1: Choose the model

For a thin lens in air, the object distance \(d_o\), image distance \(d_i\), and focal length \(f\) obey

(1)\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}.\]

We assume paraxial rays and neglect the physical thickness of the lens.

Step 2: Rearrange the equation

We need \(d_i\), so first move the object-distance term to the left:

\[\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}.\]

Put the right-hand side over a common denominator:

\[\frac{1}{d_i} = \frac{d_o}{f d_o} - \frac{f}{f d_o} = \frac{d_o-f}{f d_o}.\]

Take the reciprocal of both sides:

(2)\[d_i = \frac{f d_o}{d_o-f}.\]

Step 3: Substitute and calculate

Insert the known values into Equation (2):

\[\begin{split}\begin{aligned} d_i &= \frac{(100\ \mathrm{mm})(300\ \mathrm{mm})} {300\ \mathrm{mm}-100\ \mathrm{mm}} \\ &= \frac{30\,000\ \mathrm{mm^2}}{200\ \mathrm{mm}} \\ &= 150\ \mathrm{mm}. \end{aligned}\end{split}\]

One power of millimetres cancels, leaving the required unit of length.

Step 4: Interpret the result

The image distance is positive. Under our convention, this means the rays meet on the far side of the lens and form a real image. The image is \(150\ \mathrm{mm}\) behind the lens.

Check 1: Substitute back

Substitute \(d_i=150\ \mathrm{mm}\) into Equation (1):

\[\frac{1}{300\ \mathrm{mm}} + \frac{1}{150\ \mathrm{mm}} = \frac{1+2}{300\ \mathrm{mm}} = \frac{1}{100\ \mathrm{mm}} = \frac{1}{f}.\]

The two sides agree.

Check 2: Estimate physically

The object is at \(3f\). A real object beyond \(2f\) should produce a real image between \(f\) and \(2f\). Our result satisfies

\[100\ \mathrm{mm} < 150\ \mathrm{mm} < 200\ \mathrm{mm},\]

so its position is physically reasonable.

Final answer

\[\boxed{d_i = +150\ \mathrm{mm}}\]

The image is real and forms \(150\ \mathrm{mm}\) behind the lens.

Try it yourself

Repeat the calculation with the same lens and \(d_o=150\ \mathrm{mm}\). Before calculating, predict whether the image will be closer to or farther from the lens than the object. Then use Equation (2) and verify your result by substitution.