Lenses: Ray Diagrams and Magnification - Worksheets, Questions and Revision

11 original exam-style questions - 4 pages of questions with a full mark scheme - free printable PDF.

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This topic is chapter 1 of IGCSE Physics Practice Book.

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3.1 Lenses: Ray Diagrams and Magnification

EDEXCEL 4PH1 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

State two everyday uses of a converging lens.

(Total for Question 1 is 2 marks)

Define the principal focus of a converging lens.

(Total for Question 2 is 1 mark)

Define the focal length of a lens.

(Total for Question 3 is 1 mark)

A converging lens forms an image at v = 10 cm from an object placed at u = 4 cm from the lens. Calculate the magnification of the lens in this instance.

(Total for Question 4 is 2 marks)

State one everyday use of a diverging lens, and briefly explain why a diverging lens is suitable for it.

(Total for Question 5 is 2 marks)

For each object position given below, state the nature (real or virtual), orientation (upright or inverted) and size (magnified, diminished, or the same size) of the image formed by a converging lens.

(a)The object is beyond 2F (further from the lens than twice the focal length).(2)

(b)The object is exactly at 2F.(2)

(c)The object is between F and 2F.(2)

(d)The object is between the lens and F (closer to the lens than the focal length).(2)

(Total for Question 6 is 8 marks)

An object of height 3 cm is placed in front of a lens that produces a magnification of 5. Calculate the height of the image formed.

(Total for Question 7 is 2 marks)

A stamp collector uses a magnifying glass, which is a converging lens of focal length 5 cm, to examine a stamp. The stamp is held 3 cm from the lens.

(a)State whether the object distance (3 cm) is less than, equal to, or greater than the focal length (5 cm).(1)

(b)Hence state the nature, orientation and size of the image formed.(2)

(c)The image seen through the magnifying glass appears to be 4.5 cm tall, and the stamp itself is 1.5 cm tall. Calculate the magnification.(2)

(Total for Question 8 is 5 marks)

State the type of image (in terms of nature, orientation and size) that is ALWAYS formed by a diverging (concave) lens, regardless of the object's position, and explain why this is the case.

Figure: A diverging lens (drawn as an inward-pointing double-headed-arrow symbol) on a horizontal principal axis. An upright object arrow stands to the left. A ray parallel to the axis strikes the lens and refracts outward (away from the axis); tracing this refracted ray backwards (shown as a dashed line) meets the near-side principal focus. A second ray through the centre of the lens continues straight. The two rays diverge on the right but their backward extensions (dashed) cross on the same side as the object, forming a small upright virtual image there.

(Total for Question 9 is 3 marks)

For a certain converging lens set-up, the object distance is u = 12 cm and the image distance is v = 36 cm. Calculate the magnification, using magnification = v / u, and state whether the image is magnified or diminished compared with the object.

(Total for Question 10 is 3 marks)

A camera uses a converging lens to form a real image of a person on its sensor. The person is 1.8 m tall and stands 3.6 m from the lens. The image formed on the sensor is 2 cm tall.

(a)Convert the person's height to centimetres, and calculate the magnification of the camera lens in this instance.(2)

(b)Explain why this magnification is much less than 1, referring to the position of the person relative to the lens.(2)

(Total for Question 11 is 4 marks)

Mark scheme · 3.1 Lenses: Ray Diagrams and Magnification

Question 1

B1 any one valid use, e.g. magnifying glass
B1 any second valid use, different from the first, e.g. camera lens, projector lens, telescope objective lens, spectacle lens for long-sightedness, or the lens in the human eye
Answer: e.g. magnifying glass and camera lens (accept any two valid distinct uses)

Question 2

B1 the point at which rays travelling parallel to the principal axis converge (meet) after passing through (refracting at) the lens
Answer: The point where parallel rays converge after refracting through the lens

Question 3

B1 the distance from the centre of the lens to the principal focus
Answer: The distance from the centre of the lens to the principal focus

Question 4

M1 magnification = v / u = 10 / 4 set up
A1 magnification = 2.5 cao
Answer: Magnification = 2.5

Question 5

B1 a valid use, e.g. spectacle lenses to correct short-sightedness (myopia)
B1 valid brief reason, e.g. in a short-sighted eye the eye focuses light too strongly (in front of the retina); a diverging lens spreads the light out slightly before it reaches the eye, reducing the eye's excess focusing power so the image forms correctly on the retina
Answer: e.g. correcting short-sightedness, because it diverges light before it enters an eye that focuses too strongly

Question 6

(a) B1 real and inverted
(a) B1 diminished (smaller than the object)
(a) Answer: Real, inverted, diminished
(b) B1 real and inverted
(b) B1 the same size as the object
(b) Answer: Real, inverted, same size
(c) B1 real and inverted
(c) B1 magnified (larger than the object)
(c) Answer: Real, inverted, magnified
(d) B1 virtual and upright
(d) B1 magnified (larger than the object)
(d) Answer: Virtual, upright, magnified

Question 7

M1 rearranges magnification = image height / object height to image height = magnification x object height
A1 image height = 15 cm cao (5 x 3)
Answer: Image height = 15 cm

Question 8

(a) B1 less than (the object is inside the focal length)
(a) Answer: Less than (3 cm < 5 cm)
(b) B1 virtual and upright
(b) B1 magnified, ft from part (a)
(b) Answer: Virtual, upright, magnified
(c) M1 magnification = image height / object height = 4.5 / 1.5 set up
(c) A1 magnification = 3 cao
(c) Answer: Magnification = 3

Question 9

B1 virtual
B1 upright and diminished (smaller than the object), formed between the lens and its principal focus
B1 correct reason: a diverging lens spreads out (diverges) rays that were parallel, so the refracted rays never actually meet on the far side; they only appear to come from a point on the near side when traced backwards
Answer: Always virtual, upright, diminished

Question 10

M1 magnification = v / u = 36 / 12 set up
A1 magnification = 3 cao
B1 magnified, since magnification is greater than 1, ft their value
Answer: Magnification = 3; the image is magnified

Question 11

(a) M1 converts 1.8 m to 180 cm, and forms magnification = image height / object height = 2 / 180
(a) A1 magnification = 1/90 oe (awrt 0.0111), cao
(a) Answer: Magnification = 1/90 (approximately 0.0111)
(b) B1 the person is much further from the lens than 2F (camera lenses have a short focal length compared with typical subject distances)
(b) B1 so, as in Question 5(a), the image formed is real, inverted and greatly diminished, consistent with a magnification far below 1
(b) Answer: The person is far beyond 2F, so the image is real, inverted and greatly diminished

Lenses: Ray Diagrams and Magnification - Worksheets, Questions and Revision

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3.1 Lenses: Ray Diagrams and Magnification

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