Fig. 43.1 shows a simplified diagram of a digital camera. A converging lens forms an image of a distant building on an electronic sensor at the back of the ...
Question 1 Report
Fig. 43.1 shows a simplified diagram of a digital camera. A converging lens forms an image of a distant building on an electronic sensor at the back of the camera body. The focal length of the lens is 50 mm. The building is 18 m tall and stands 300 m from the camera. A photographer adjusts the focus ring until the image on the sensor is sharp. The sensor measures 24 mm × 36 mm. Light enters through the aperture and is focused by the lens onto the sensor plane. The camera body is enclosed so that no stray light reaches the sensor.
(a) State why the image distance is approximately equal to the focal length when the object is far away. [1]
(b) Calculate the height of the image of the building on the sensor. [2]
(c) Describe three properties of the image formed on the sensor. [2]
(d) Describe how the image would change if the photographer replaced the lens with one of focal length 100 mm. [2]
Answer Details
Marking Scheme
- (a) [1] When the object distance u is very large, 1/u is approximately zero, so from 1/f = 1/u + 1/v, v is approximately equal to f.
- (b) [1] Magnification = v/u approximately equals f/u = 50/300,000 = 1.67 x 10-4. [1] Image height = 1.67 x 10-4 x 18,000 = 3.0 mm.
- (c) [1] Real. [1] Inverted and diminished.
- (d) [1] The image would be larger (doubled to about 6.0 mm) because magnification increases with focal length. [1] The field of view would be narrower.
Explanation
(a) The thin lens equation is 1/f = 1/u + 1/v. When the object is very far away (300 m = 300,000 mm), 1/u = 1/300,000 is extremely small compared to 1/f = 1/50. Therefore 1/v is approximately equal to 1/f, meaning v is approximately equal to f. Physically, rays from a very distant object arrive at the lens nearly parallel, and parallel rays converge at the focal point.
(b) All distances must be in the same unit. Converting to mm: u = 300 m = 300,000 mm, f = 50 mm, object height = 18 m = 18,000 mm.
Since v is approximately equal to f:
magnification = v/u = 50/300,000 = 1.667 x 10-4
image height = magnification x object height = 1.667 x 10-4 x 18,000 = 3.0 mm
The 18 m tall building produces a tiny 3.0 mm image on the sensor.
(c) The image is real because it forms on the opposite side of the lens from the object, where actual light rays converge. It can be captured on the sensor (or a screen). The image is inverted (upside down compared to the object) and diminished (much smaller than the object), both of which are characteristic of a real image formed when the object is well beyond 2f.
(d) Doubling the focal length from 50 mm to 100 mm doubles the magnification (since magnification is approximately f/u for distant objects). The image height would become about 6.0 mm. However, a longer focal length lens captures a narrower field of view because it magnifies a smaller angular portion of the scene. This is the principle behind telephoto lenses in photography: longer focal length gives greater magnification but a narrower view.
Download The App On Google Playstore
Everything you need to excel in your exams
