Matching Faded Lens Labels to Three Optics Jobs
In a school optics tray, three lenses are marked +4.0 D, +10.0 D and −2.0 D, but their labels have faded. One student has myopia and can read a book at 25 cm but cannot see the board clearly. Another student wants a magnifying glass for reading small print held about 6 cm from the lens. The teacher also asks for a lens that will form a real image of a candle on a screen placed 40 cm from the lens. The students must match each lens to a use, calculate focal lengths, and check whether the magnifier at 6 cm gives a virtual, erect image. They are warned that a higher power is not automatically better for every job and that the bench is only 80 cm long.
1. Which lens should correct the myopic student's distance vision? Calculate its focal length and state the image character it produces for a distant object.
Read the worked answer
Myopia is corrected with a diverging lens, so the choice is the negative-power lens, −2.0 D. Focal length f = 1/P = 1/(−2.0 m⁻¹) = −0.50 m = −50 cm. For a distant object, a concave lens forms a virtual, erect, diminished image on the same side as the object. The eye's lens is too strong for the eyeball length, so the diverging lens spreads the rays slightly and moves the focused image back onto the retina. The +4.0 D and +10.0 D lenses are converging and would worsen myopia.
2. Select the magnifier and calculate the image position and magnification when the object is 6 cm from it. Is the image virtual and erect?
Read the worked answer
The +10.0 D lens has f = 0.10 m = 10 cm. Since 6 cm is inside the focal length, it should act as a magnifier. Using 1/v − 1/u = 1/f with u = −6 cm and f = +10 cm: 1/v = 1/10 + 1/(−6) = 0.100 − 0.1667 = −0.0667, so v = −15 cm. The negative image distance means a virtual image on the object side. Magnification m = v/u = (−15)/(−6) = +2.5, so the image is erect and 2.5 times larger. The +4.0 D lens (f = 25 cm) would also give a virtual image but with much less magnification at 6 cm.
3. For the projection experiment, which lens should be used and where should the candle be placed? Why are the other choices unsuitable?
Read the worked answer
A real image on a screen requires a converging lens, so the −2.0 D lens is out. For the +10.0 D lens, f = 10 cm and v = +40 cm. Using 1/u = 1/v − 1/f = 1/40 − 1/10 = 0.025 − 0.100 = −0.075, so u = −13.3 cm. Magnification is v/u = 40/(−13.3) = −3, giving an inverted real image. The +4.0 D lens would need u = −66.7 cm for the same screen distance, so object-to-screen distance would be about 106.7 cm, exceeding the 80 cm bench. Therefore the +10.0 D lens is the practical choice.