A converging lens f=20cm — object at 30cm — find image position and magnification

Question

An object is placed 30 cm in front of a converging (convex) lens of focal length 20 cm. Find the position of the image and the magnification produced.

Solution — Step by Step

Identify given values and sign convention

Using the standard sign convention (all distances measured from optical centre; distances in the direction of incident light are positive):

Object distance: $u = -30$ cm (object is on the left of lens, so negative)
Focal length: $f = +20$ cm (converging lens, so positive)
Image distance: $v = ?$

Apply the lens formula

The lens formula is:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Substituting:

$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20}$

$\frac{1}{v} + \frac{1}{30} = \frac{1}{20}$

$\frac{1}{v} = \frac{1}{20} - \frac{1}{30}$

Solve for v

Finding a common denominator (LCM of 20 and 30 is 60):

$\frac{1}{v} = \frac{3}{60} - \frac{2}{60} = \frac{1}{60}$

$\boxed{v = +60\text{ cm}}$

The positive sign means the image is formed on the opposite side of the lens from the object — this is a real image.

Calculate magnification

Linear magnification:

$m = \frac{v}{u} = \frac{+60}{-30} = -2$

Interpretation: The magnitude 2 means the image is twice the size of the object. The negative sign means the image is inverted (real and inverted).

Why This Works

The lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ is derived from Snell's law applied to paraxial rays through a thin lens. Each surface of the lens refracts light, and the combined effect is captured in this single formula.

The sign of $v$ tells us the image type:

$v > 0$ : real image (on the transmission side, can be captured on screen)
$v < 0$ : virtual image (on the same side as the object, cannot be on screen)

Here, the object is between $f$ and $2f$ (20 cm and 40 cm), at 30 cm. For a convex lens, an object between $f$ and $2f$ produces a real, inverted, magnified image beyond $2f$ . Our answer of 60 cm confirms this — 60 cm is beyond 40 cm ( $2f$ ).

Alternative Method — Using Cartesian Form

Some textbooks write the lens formula as $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ and others as $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ . These are identical. The key is consistent application of the sign convention. Using $u = -30$ :

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \implies \frac{1}{20} = \frac{1}{v} + \frac{1}{30}$

Same calculation, same answer.

Common Mistake

⚠️ Common Mistake

Using $u = +30$ (forgetting the negative sign). In the standard sign convention, object distance $u$ is always negative for a real object (because the object is on the same side as incoming light, which is the negative direction). Using $u = +30$ gives $\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}$ ... accidentally getting the right numerical answer but for the wrong reason. When $u$ is substituted with wrong sign into other problems, you get completely wrong results.

💡 Expert Tip

Quick check for convex lens: when object is at $2f = 40$ cm, image is also at $2f = 40$ cm and magnification = 1 (same size). Our object at 30 cm is between $f$ and $2f$ , so image should be beyond $2f$ and magnified. Our answer: $v = 60$ cm (beyond 40 cm) and $|m| = 2$ (magnified). This mental check confirms we're on the right track.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Cartesian Form

Common Mistake

Try These Next