formula.expert

Converging lens: where the image lands and how tall it is

SNC2D Grade 10 Science · Light and Geometric Optics

The optical bench at the back of the lab is a metre stick fitted with sliding clamps: a lamp box with an arrow-shaped aperture, a lens holder, and a white card that serves as the screen. A student mounts a converging lens — its focal length stamped on the rim as 12.0 cm — in the holder, then clamps the lit arrow, which measures 4.50 cm tall, at a mark 30.0 cm in front of the lens. With the room lights dimmed, the student slides the white card along the far side of the bench and watches a smear of light sharpen, snap into a crisp inverted arrow, and blur again as the card passes through focus. The card is eased back until the image is at its sharpest. Find how far the card sits from the lens, the magnification of the image, and how tall the image is.

4.50 cm arrowf = 12.0 cm30.0 cm20.0 cmimage 3.00 cm, inverted

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • h_o = 4.5 cm — Height of the lit arrow
  • d_o = 30 cm — Object distance from the lens
  • f = 12 cm — Focal length of the converging lens
Determine
  1. (a)how far from the lens the card comes into focus
  2. (b)the magnification of the image
  3. (c)how tall the image is
Step 1 of 3(a) · solve for Image distance

Where the card has to go is the first thing you can compute — nothing about the image's size is knowable until you know where it forms.

fdodi
Rearranged for d_i
di=(1f−1do)−1d_i = \left(\tfrac{1}{f} - \tfrac{1}{d_o}\right)^{-1}
Your values, in your units
di=(1(12 cm)−1(30 cm))−1d_i = \left(\tfrac{1}{\left( 12\ \text{cm} \right)} - \tfrac{1}{\left( 30\ \text{cm} \right)}\right)^{-1}
Converted to base units
di=(1(0.12 m)−1(0.3 m))−1d_i = \left(\tfrac{1}{\left( 0.12\ \text{m} \right)} - \tfrac{1}{\left( 0.3\ \text{m} \right)}\right)^{-1}
Answer
di=200 mmd_i = 200\ \text{mm}

Carried onward at full precision, not this rounded figure.

Open the Thin Lens Equation solver →

Step 2 of 3(b) · solve for Magnification

Magnification is a ratio of the two distances, so the answer from step 1 comes straight back in. The minus sign is not decoration: it is the statement that the image is upside down.

mdodi
Rearranged for m
m=−didom = -\tfrac{d_i}{d_o}
200 mmcarried from step 1
Your values, in your units
m=−(0.2 m)(30 cm)m = -\tfrac{\left( 0.2\ \text{m} \right)}{\left( 30\ \text{cm} \right)}
Converted to base units
m=−(0.2 m)(0.3 m)m = -\tfrac{\left( 0.2\ \text{m} \right)}{\left( 0.3\ \text{m} \right)}
Answer
m=−666.67 per millem = -666.67\ \text{per mille}

Carried onward at full precision, not this rounded figure.

Open the Lens Magnification (m = −d_i/d_o) solver →

Step 3 of 3(c) · solve for Image height

Same magnification, now read as a ratio of heights instead of distances. Multiply the object's 4.50 cm by it and the negative answer says the arrow on the card points down.

hohim
Rearranged for h_i
hi=m hoh_i = m \, h_o
-666.67 per millecarried from step 2
Your values, in your units
hi=(−0.666667) (4.5 cm)h_i = \left( -0.666667 \right) \, \left( 4.5\ \text{cm} \right)
Converted to base units
hi=(−0.666667) (0.045 m)h_i = \left( -0.666667 \right) \, \left( 0.045\ \text{m} \right)
Answer
hi=−30 mmh_i = -30\ \text{mm}

Carried onward at full precision, not this rounded figure.

Open the Magnification from Heights (m = h_i/h_o) solver →

Answer

Therefore the card focuses 20.0 cm behind the lens, the magnification is −2/3, and the 4.50 cm arrow lands on the card as a 3.00 cm image — reduced and, as the minus sign insists, upside down.

Why this order

The order here is forced by the physics, not by convenience. The thin lens equation contains no information about size at all — it only locates the image — so step 1 has to come first, and every later answer hangs off its 20.0 cm. Step 2 is the hinge of the whole chain: magnification is defined twice over, once as −di/do and once as hi/ho, and the reason a lens problem can be solved at all is that those two definitions describe the same number. Step 3 simply reads the ratio the other way round.

Two things go wrong reliably. The first is dropping the minus sign, which turns a real inverted image into a fictional upright one; slide a card into the beam on a real bench and the arrow is unmistakably upside down, so the sign is checkable in ten seconds. The second is treating a bigger image distance as a bigger image without checking against the object distance — here di = 20.0 cm is smaller than do = 30.0 cm, so the image is reduced, |m| = 2/3, and the 4.50 cm arrow lands as a 3.00 cm one. That is the geometry inside every camera: the sensor sits closer to the glass than the subject does, which is precisely why a whole hockey rink fits on a chip the size of a fingernail.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.