Questions on lenses and images

Question 1

State what is meant by the principal focus and the focal length of a converging lens.

Accepted Answer

The principal focus is the point where rays arriving parallel to the principal axis are brought together after passing through the lens (1). The focal length is the distance from the centre of the lens to the principal focus (1).

Question 2

Calculate the power of a converging lens of focal length 40 cm, giving the unit.

Accepted Answer

f = 40 cm = 0.40 m (1) P = 1/f = 1/0.40 = 2.5 D, dioptres (1)

Question 3

State the difference between a real image and a virtual image.

Accepted Answer

A real image is formed where rays actually meet, so it can be caught on a screen (1). A virtual image is formed where rays only appear to come from when projected backwards; no rays pass through it and no screen can catch it (1).

Question 4

An object is placed in front of a diverging lens. State the nature of the image formed.

Accepted Answer

Virtual: it forms where the spreading refracted rays appear to come from, on the same side of the lens as the object (1). It is upright and diminished (1).

Question 5

A converging lens is used to photograph a distant building. State where the image forms, and give its nature.

Accepted Answer

The image forms in the focal plane: the image distance is (approximately) equal to the focal length, because 1/u is negligible for a distant object (1). It is real, inverted and diminished (1).

Question 6

A student writes the magnification of an image as m = 0.60 m. State the error the student has made.

Accepted Answer

Magnification is a ratio of two lengths (v/u, or image height over object height), so it has no unit; the metre should not appear (1).

Question 7

An object is placed 0.45 m from a converging lens of focal length 0.15 m. Calculate the image distance and the magnification.

Accepted Answer

1/v = 1/f − 1/u = 1/0.15 − 1/0.45 (1) v = 0.225 m ≈ 0.23 m (1) m = v/u = 0.50 (1)

Question 8

Describe the paths of the three standard construction rays used to locate the image formed by a thin converging lens.

Accepted Answer

A ray parallel to the principal axis refracts to pass through the principal focus on the far side (1). A ray through the centre of the lens continues undeviated (1). A ray passing through the near principal focus leaves the lens parallel to the axis (1). Any two fix the image; the third confirms it.

Question 9

An object is placed 0.10 m from a converging lens of focal length 0.25 m. Calculate the image position, state the nature of the image, and give its magnification.

Accepted Answer

1/v = 1/0.25 − 1/0.10 = −6.0 m⁻¹ (1)
v = −0.167 m ≈ −0.17 m (1)
The negative sign means the image is virtual: upright and on the same side as the object (1)
m = |v|/u = 1.67, so the image is magnified (1)

Question 10

A student covers the top half of a lens with card and predicts the top half of the image will vanish. State and explain what actually happens.

Accepted Answer

The whole image remains, but dimmer (1). Every point of the object sends rays through every part of the lens, so the uncovered half still forms the complete image with half the light (1).

Question 11

A diverging lens has a power of −4.0 D. Calculate its focal length.

Accepted Answer

f = 1/P = 1/(−4.0) = −0.25 m (1). The negative sign marks the lens as diverging: parallel rays spread as if from a focus 25 cm on the incoming side (1).

Question 12

An object is placed 60 cm from a converging lens of power +2.5 D. Calculate the image distance and the magnification.

Accepted Answer

u = 60 cm = 0.60 m, and 1/f = P, so 1/v = P − 1/u = 2.5 − 1/0.60 (1). v = 1.2 m (1). m = v/u = 1.2/0.60 = 2.0 (1).

Question 13

A converging lens of focal length 0.15 m forms a real image three times the height of the object. Calculate the object distance and the image distance.

Accepted Answer

m = v/u = 3, so v = 3u (1). Substituting into the lens equation: 1/u + 1/(3u) = 1/0.15, so (4/3)(1/u) = 1/0.15 (1). u = 0.20 m (1). v = 3u = 0.60 m (1).

Question 14

An object 2.0 cm tall stands 0.80 m from a converging lens of focal length 0.25 m. Calculate the height of the image and state its orientation.

Accepted Answer

1/v = 1/f − 1/u = 1/0.25 − 1/0.80 = 2.75 m⁻¹, so v = 0.364 m (1). m = v/u = 0.364/0.80 = 0.455 (1). Image height = 0.455 × 2.0 cm = 0.91 cm; the image is real and inverted (1).

Question 15

A lens forms a real image on a screen 0.60 m away when the object is 0.30 m from the lens. Calculate the focal length of the lens.

Accepted Answer

1/f = 1/u + 1/v = 1/0.30 + 1/0.60 (1) 1/f = 5.0 m⁻¹ (1) f = 0.20 m, that is 20 cm (1)

Question 16

A +4.0 D converging lens is placed in contact with a −1.5 D diverging lens. Calculate the power and focal length of the combination.

Accepted Answer

Powers of thin lenses in contact add (1) P = 4.0 + (−1.5) = 2.5 D (1) f = 1/P = 1/2.5 = 0.40 m, converging (1)

Question 17

A projector lens of focal length 5.0 cm throws a sharp image on a screen 3.0 m away. Calculate the object distance, and the magnification of the image on the screen.

Accepted Answer

1/u = 1/f − 1/v = 1/0.050 − 1/3.0 (1) 1/u = 19.67 m⁻¹ (1) u = 0.0508 m ≈ 5.1 cm (1) m = v/u = 3.0/0.0508 ≈ 59 (1)

Question 18

Explain, in terms of the images they form, why a magnifying glass must hold the object inside the focal length, and what happens to the image as the object slides out towards the focal point.

Accepted Answer

Inside the focal length the refracted rays diverge, so only their backward projections meet: the image is virtual, upright and magnified, which is what a magnifying glass is for (1). As the object approaches the focal point the projections meet further and further away and the image grows (1); at the focal point itself the rays emerge parallel and the image forms at infinity (1).

Question 19

A camera has a converging lens of focal length 50 mm which moves towards or away from the sensor to focus. Calculate how far, and in which direction, the lens must move from its position when focused on a very distant object so that an object 1.2 m away is in focus.

Accepted Answer

For a very distant object 1/u ≈ 0, so the sensor sits at v = f = 50 mm = 0.050 m (1). For u = 1.2 m: 1/v = 1/0.050 − 1/1.2 = 19.2 m⁻¹ (1), so v = 52.2 mm (1). The lens must move 2.2 mm away from the sensor (1).

Question 20

A student must project the image of a slide onto a screen fixed 2.4 m from the lens, with the image at least 20 times the height of the slide. Two converging lenses are available:
Lens L₁: focal length 0.10 m.
Lens L₂: focal length 0.25 m.
Deduce which lens the student should use, and state the distance from the slide to the lens.

Accepted Answer

For both lenses v = 2.4 m. L₁: 1/u = 1/0.10 − 1/2.4 (1), so u = 0.104 m and m = v/u = 2.4/0.104 = 23 (1). L₂: u = 0.279 m, so m = 8.6 (1). Only L₁ gives a magnification of at least 20, so the student should use L₁ (1), with the slide 10.4 cm from the lens (1).

Question 21

Describe an experiment to determine the focal length of a converging lens, using an illuminated object, the lens, a screen and a metre rule. Include the measurements taken, how the focal length is found from them, and one way to check or improve the result.

Accepted Answer

Line up the illuminated object (for example a lamp behind crosswires), the lens and the screen, and move the screen until the image is sharp (1). Measure the object distance u and the image distance v from the centre of the lens with the metre rule (1). Repeat for a range of object distances (1). Calculate 1/u and 1/v for each pair and plot 1/v against 1/u (or find f from 1/f = 1/u + 1/v for each pair and average) (1). The graph is a straight line whose intercept on each axis equals 1/f, giving the focal length (1). Check or improve: focus a distant object (a window across the room) onto the screen, where the image distance is approximately f; or approach the sharpest image position from both sides and average (1).