Slide Projector: Image Size Limits in a Small Room
A student is setting up a slide projector using a single converging lens of focal length 15.0 cm. With the screen 1.20 m from the lens he obtains a sharp image of a slide 3.5 cm wide. He is then asked to produce an image 0.90 m wide for a display. The greatest distance available between the lens and the screen in the room is 3.00 m. He moves the screen to the 3.00 m mark, refocuses by moving the lens, and finds the image is still only about 66 cm wide, and noticeably dimmer than before. He also notices that the slide had to be loaded upside down and back to front. He wonders whether a different lens would solve the problem, and whether the dimmer image is a fault in the lamp.
1. For the screen at 1.20 m, calculate the object distance and the width of the image.
Read the worked answer
Use 1/f = 1/u + 1/v with f = 15.0 cm and v = 120 cm:
1/u = 1/15 - 1/120 = 8/120 - 1/120 = 7/120
u = 120/7 = 17.1 cm (the slide sits 17.1 cm from the lens)
Magnification m = v/u = 120 / 17.1 = 7.0
Image width = 7.0 x 3.5 cm = 24.5 cm
The image is real, inverted and magnified. Note that u is only just greater than f; that is why the magnification is large.
2. Can he produce an image 0.90 m wide with this lens in a room where the lens-to-screen distance cannot exceed 3.00 m? What would be needed instead?
Read the worked answer
Required magnification = 0.90 m / 0.035 m = 25.7.
With this lens, the largest magnification available is at v = 300 cm:
1/u = 1/15 - 1/300 = 20/300 - 1/300 = 19/300, so u = 15.8 cm and m = 300 / 15.8 = 19.0
Largest image = 19.0 x 3.5 cm = 66.5 cm. So 0.90 m is impossible here - the lens is too long in focal length and the room too short.
What is needed: for m = 25.7 with v = 300 cm, u = 300 / 25.7 = 11.7 cm, and
1/f = 1/11.7 + 1/300 = 0.0855 + 0.0033 = 0.0888
f = 11.3 cm. A lens of about 11 cm focal length would give the required size. Equivalently he could keep this lens and move the screen beyond 4 m, which the room does not allow.
3. Explain why the slide must be loaded upside down and back to front, and why the image is dimmer at the larger size.
Read the worked answer
A single converging lens forms a real image where the rays actually cross; after crossing, the top of the object is at the bottom of the image and left is on the right. So the slide must be inverted and reversed for the audience to see it the right way up.
Brightness: the lamp delivers roughly the same luminous flux regardless of magnification, but the image area grows as m^2. Going from m = 7.0 to m = 19.0 increases the area by (19/7)^2 = 7.4 times, so the illuminance falls by roughly that factor. The image is dimmer because the same energy is spread over more area - not because the lamp has developed a fault. Some light also misses the lens aperture or falls outside the screen, which makes the loss slightly worse than the area ratio alone predicts.