Sunita's convex lens: why the screen image vanishes
Sunita's physics teacher hands her a convex lens and tells her only that its focal length is 15 cm. Her task is to form a real image of a candle flame on a screen using the school optical bench. She places the candle 20 cm from the lens and, after some adjustment, finds a sharp image on a screen 60 cm beyond the lens. Encouraged, she moves the candle to 10 cm from the lens and searches for a screen image for five minutes; nothing appears wherever she puts the screen. She concludes that the lens is damaged. Her partner disagrees and says the lens is fine - she is simply looking for the wrong kind of image in the wrong place. Later the teacher asks them to use the same lens to throw a picture of a small transparency onto the classroom wall, which will stand 90 cm from the lens.
1. Verify Sunita's first result using the lens formula and state the nature, position and magnification of the image she obtained with the candle at 20 cm.
Read the worked answer
Using the sign convention with distances measured from the optical centre, object distance u = −20 cm and focal length f = +15 cm for a convex lens:
1/v − 1/u = 1/f 1/v = 1/f + 1/u = 1/15 + (−1/20) = (4 − 3)/60 = 1/60 v = +60 cm
This matches the position where she found the sharp image, so the observation is consistent and the lens is behaving normally.
Magnification m = v/u = 60/(−20) = −3.
The negative sign means the image is inverted, and |m| = 3 means it is three times the size of the flame. Since v is positive, the image is real and can be caught on a screen. This is exactly the arrangement of a projector or a cinema lens: object between f and 2f, image beyond 2f, inverted and magnified.
2. Explain why no image appears on the screen when the candle is 10 cm from the lens, and state where the image actually is and how Sunita should look at it.
Read the worked answer
At u = −10 cm the object lies inside the focal length (10 cm < 15 cm). Rays from the flame strike the lens and, after refraction, still diverge on the other side. Diverging rays cannot meet, so no real image can be formed and no screen position will ever show a sharp picture. The lens is not damaged.
The lens formula confirms this: 1/v = 1/15 + (−1/10) = (2 − 3)/30 = −1/30, so v = −30 cm.
The negative v means the image is formed on the same side as the object, 30 cm from the lens. m = v/u = (−30)/(−10) = +3: the image is virtual, erect and three times magnified.
A virtual image can only be seen by the eye, not caught on a screen. Sunita should remove the screen, put her eye close to the lens on the far side, and look back through the lens at the candle. She will see an enlarged, upright flame apparently located behind the lens. This is precisely how a magnifying glass works, and it is why a magnifying glass cannot project an image onto a wall.
3. For the transparency to be imaged on the wall 90 cm from the lens, where must the transparency be placed, and how should it be oriented? Show the calculation.
Read the worked answer
Here the image must be real and on the far side of the lens, so v = +90 cm and f = +15 cm. Find u:
1/v − 1/u = 1/f 1/u = 1/v − 1/f = 1/90 − 1/15 = (1 − 6)/90 = −5/90 1/u = −1/18, so u = −18 cm
The transparency must be held 18 cm in front of the lens - that is, between f (15 cm) and 2f (30 cm), which is the correct zone for producing a magnified real image.
Magnification m = v/u = 90/(−18) = −5. The image is five times enlarged and inverted.
Because the wall image is inverted, the transparency must be placed upside down (and, if it carries text, reversed left-to-right as well) so that it appears the right way up to the class. This is exactly why slides in an old slide projector were inserted inverted. A practical warning: at 18 cm the image is 5× larger, so the wall must be big enough, and a small error in the 18 cm setting throws the image badly out of focus because the magnification is large.