The Concave Mirror That Would Not Focus
In a school laboratory a student must project a magnified image of a small candle flame onto a white card using a concave mirror of focal length 15 cm. He sets the candle 10 cm in front of the mirror and searches for the image with the card at distances between 20 cm and 40 cm in front of the mirror. No image appears on the card at any position, although when he looks into the mirror from a metre away he clearly sees a large, erect flame. He concludes that the mirror is defective and asks for a plane mirror. The teacher checks the mirror against a distant window and finds it forms a sharp image at 15 cm, so the mirror is sound. The card is fine and the room is dim enough for a bright image. The optical bench allows the candle to be placed anywhere from 5 cm to 80 cm from the mirror.
1. Use the mirror formula to show why no image can be caught on the card, and account for the large erect flame the student sees.
Read the worked answer
Take the pole as origin, distances towards the incident light as negative: u = −10 cm, f = −15 cm.
1/v + 1/u = 1/f, so 1/v = 1/f − 1/u = −1/15 − (−1/10) = −1/15 + 1/10 = (−2 + 3)/30 = +1/30, giving v = +30 cm.
A positive v means the image lies behind the mirror. Reflected rays diverge as if from that point; they never actually meet, so no card anywhere can intercept a real image. Magnification m = −v/u = −(+30)/(−10) = +3: erect, three times enlarged, 30 cm behind the mirror.
That is precisely the large erect flame he sees when looking into the mirror from a metre away - he is seeing the virtual image with his eye, not a failure of the apparatus. A candle inside the focal length always gives a virtual, erect, magnified image; the fault is the object distance, not the mirror.
2. State where the candle must be placed for a real image, and give one complete working setting: candle position, card position and magnification.
Read the worked answer
A real image requires the object beyond the focal point, i.e. more than 15 cm from the mirror. The student's search range of 20–40 cm for the card also failed partly because his object was wrong; the real image position must be calculated, not guessed.
Take u = −20 cm: 1/v = −1/15 + 1/20 = (−4 + 3)/60 = −1/60, so v = −60 cm.
The negative sign means the image is 60 cm in front of the mirror, on the same side as the candle - a real, inverted image. m = −v/u = −(−60)/(−20) = −3, so it is three times enlarged and inverted. The card must go at 60 cm, well beyond the range he was searching.
Any object distance between 15 cm and 30 cm gives a magnified real image; 25 cm would give v = −37.5 cm with m = −1.5, a gentler magnification that is easier to focus on the bench.
3. He now covers the upper half of the mirror with black paper and keeps the candle at 20 cm. Predict exactly what appears on the card at 60 cm, and justify your prediction.
Read the worked answer
A complete, correctly positioned image still appears at 60 cm - three times enlarged and inverted - but it is fainter, roughly half as bright.
Reasoning: every point of the flame sends light to every part of the mirror, not just to a matching point. The uncovered lower half alone still receives rays from the tip and from the base of the flame and reflects them according to the same law, so all the image points are still formed in their original places. The geometry of image formation depends on the curvature of the reflecting surface, and the remaining half has exactly the same curvature and focal length.
What changes is the light-gathering area: half the aperture collects roughly half the rays, so the image is dimmer. The common wrong answer - 'half the image disappears' - confuses object points with mirror area.