Convex lens with the object inside the focal length
A student is given a convex lens and told its focal length is about 15 cm. With an object pin 30.0 cm from the lens, a sharp inverted image is found on a screen 30.0 cm from the lens. The pin is then moved to 10.0 cm from the lens. The student searches for an image on the screen on both sides of the lens but finds none, and concludes that the lens has lost its power or that the lens formula fails when the object is inside the focal length. Looking through the lens from the far side, however, the student sees an upright, magnified image of the pin. A plane mirror, a small screen, a ruler, and a second pin are available.
1. Use the first measurement to find f, then predict what happens at u = 10.0 cm. Is the student's conclusion correct?
Read the worked answer
From the first measurement, 1/f = 1/30.0 + 1/30.0 = 1/15.0, so f = 15.0 cm. At u = 10.0 cm, 1/v = 1/f - 1/u = 1/15.0 - 1/10.0 = -1/30.0, so v = -30.0 cm. The negative sign means the image is virtual and on the same side as the object. The magnification is m = -v/u = +3, so the image is upright and three times larger. The lens formula works; the lens has not lost power. The student's mistake is expecting a virtual image to be projectable on a screen.
2. Explain why the screen method fails and describe how to locate the virtual image with the available apparatus.
Read the worked answer
A screen shows an image only where light rays actually converge. For u < f, the rays leave the lens diverging as if from a point on the object side, so no real image forms on either side and no screen position will be sharp. The virtual image can be located by parallax: place the second pin on the far side of the lens and view it and the image of the first pin through the lens, moving the second pin until the two show no relative shift when the eye moves sideways. The second pin is then at the virtual image position. A plane mirror can also be used to convert the virtual image into a real one for a screen.