Projecting a Slide: Where the Screen Must Go
In the school physics laboratory a convex lens of focal length 15 cm is clamped on an optical bench. A slide 3.0 cm tall is held on the left of the lens, and a white screen stands on a bracket 40 cm to the right of the lens. The bench is 1.20 m long, so the lens, the slide and the screen can each be slid and re-clamped. A student adjusts the slide position until a sharp image of the slide appears on the screen. The teacher then asks for two changes: (i) the slide is moved to 20 cm from the lens and the screen is re-positioned for a sharp image; (ii) the slide is moved to 10 cm from the lens and the student is told to find the sharp image on the same screen. The student uses 1/v − 1/u = 1/f with all distances measured from the lens and reports magnification as m = v/u. The room lights are switched off throughout.
1. For the first position, find the distance of the slide from the lens and the height and nature of the image on the screen.
Read the worked answer
Take the direction of incident light (left to right) as positive, so f = +15 cm and the screen gives a real image at v = +40 cm.
1/v − 1/u = 1/f 1/40 − 1/u = 1/15 1/u = 1/40 − 1/15 = 3/120 − 8/120 = −5/120 = −1/24 u = −24 cm, i.e. the slide is 24 cm to the left of the lens.
Magnification m = v/u = 40/(−24) = −1.67 Image height = 1.67 × 3.0 = 5.0 cm
The negative sign means the image is real, inverted and magnified, formed on the far side of the lens where the screen is. The bench is long enough: 24 cm + 40 cm = 64 cm of the 120 cm is used.
2. The slide is moved to 20 cm from the lens. Find the new image distance and magnification, and state how far and in which direction the screen must be moved.
Read the worked answer
Now u = −20 cm, f = +15 cm.
1/v = 1/f + 1/u = 1/15 − 1/20 = 4/60 − 3/60 = 1/60 v = +60 cm
m = v/u = 60/(−20) = −3, so the image is 9.0 cm tall, real and inverted.
The slide moved closer to the lens (from 24 cm to 20 cm) and the image moved further away (from 40 cm to 60 cm). The screen must be moved 20 cm away from the lens, to 60 cm on the right. Total bench used = 20 + 60 = 80 cm, still within 120 cm.
The reasoning to state explicitly: for a converging lens, moving the object towards the focal point always pushes a real image further out and increases magnification - which is why a projector needs a long throw for a big picture.
3. The slide is now placed 10 cm from the lens. The student is asked to obtain a sharp image on the screen. What should the student report, and why is this not a failure of the lens formula?
Read the worked answer
u = −10 cm, f = +15 cm. Since the object is inside the focal length, predict first that no real image can be formed on a screen.
1/v = 1/15 − 1/10 = 2/30 − 3/30 = −1/30 v = −30 cm
m = v/u = (−30)/(−10) = +3
The image is virtual, erect, magnified (9.0 cm) and located 30 cm on the same side as the slide. Light rays leaving the lens diverge, so they never meet; a screen placed anywhere on the right will show only a blurred patch.
What the student should report: no image can be caught on the screen at this position. To see the image, the observer must look through the lens from the right side - the arrangement is a magnifying glass.
The lens formula has not failed: it returned a negative v, and the sign convention defines negative v as a virtual image on the object side. The error to avoid is treating v = −30 cm as '30 cm on the right' and sliding the screen there, or reporting 'no solution'.