Virtual Image from a Projector Lens
A student is asked to project the image of a small illuminated slide onto a screen using a converging lens of focal length 15 cm. The slide is placed 10 cm from the lens on the principal axis. The student sets a screen 30 cm on the opposite side of the lens, observes only a blurred patch, and concludes that the lens is defective. A lab assistant checks the bench: the lens is clean, the room is darkened, and the slide and screen are both perpendicular to the axis. The student has not moved the screen to search for a sharp image. The assistant asks for a ray-diagram-style calculation using the thin-lens sign convention, a decision about whether a real image can be formed with this object distance, and a corrected arrangement that would give a real image magnified by 3. The lens is thin and paraxial; the screen is opaque and cannot be placed on the same side as the object.
1. Using the Cartesian sign convention, calculate the image distance and magnification for the 10 cm object distance. Is the image real or virtual, and where does it lie relative to the lens?
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Use Cartesian sign convention with light travelling left to right: f = +15 cm, u = −10 cm. The thin-lens equation is 1/v − 1/u = 1/f. Thus 1/v = 1/f + 1/u = 1/15 − 1/10 = (2 − 3)/30 = −1/30, so v = −30 cm.
The negative image distance means the image is on the same side as the object: it is virtual. Magnification m = v/u = (−30)/(−10) = +3. The image is erect and three times the object size. Because the object is inside the focal length (10 cm < 15 cm), the lens acts as a magnifying glass, not as a projector.
2. Explain why the screen at 30 cm on the opposite side cannot show a sharp image, and why moving the screen to the virtual image position on the object side would not help either.
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A screen can show only a real image, where rays actually converge and then diverge from a point on the screen. Here the outgoing rays diverge as if they came from a point 30 cm on the object side; they never meet on the opposite side. Placing the screen at +30 cm therefore intercepts a diverging beam and gives only a blurred patch.
Moving the screen to the virtual image position on the object side would not help either: the rays are still diverging when they reach that plane, so no sharp real image forms there. The lens is not defective; the object is simply too close.
3. Find the object distance and screen distance that give a real image magnified by 3. Confirm that both lie outside the focal length in the required way.
Read the worked answer
For a real image, v is positive and m = v/u = −3 (inverted). Let the object distance be d, so u = −d and v = +3d. Substituting in 1/v − 1/u = 1/f: 1/(3d) + 1/d = 1/f, so 4/(3d) = 1/15, giving 3d = 60 and d = 20 cm. Then v = 3d = 60 cm.
The slide must be 20 cm from the lens and the screen 60 cm on the opposite side. This is outside the focal length on both counts: the object is beyond f (20 cm > 15 cm), and the image is real. The magnification is 3 and the image is inverted.