An object is placed 10 cm in front of a converging lens with focal length 15 cm. Which correctly describes the image?
AReal, inverted, on the far side of the lens
BVirtual, upright, on the same side as the object
CReal, upright, on the far side of the lens
DNo image forms because the object is inside the focal length
When s_o < f, solving 1/s_i = 1/f − 1/s_o gives a negative s_i (here, −30 cm), meaning the image is virtual and on the same side as the object. The magnitude |m| = 30/10 = 3 and m is positive — upright and magnified. This is exactly the magnifying-glass case. Option D is wrong: an image always forms; it is simply virtual when s_o < f.
Question 2 Multiple Choice
A slide projector places a slide just outside the focal length of a converging lens (s_o slightly > f). The image on the distant screen is inverted, so the slide must be loaded upside-down. Which statement best explains why the image is inverted?
AThe lens flips the image because it is a diverging lens
BWhen s_o > f for a converging lens, s_i > 0 (real image on the far side), and the magnification m = −s_i/s_o is negative, indicating an inverted image
CThe image is inverted only because s_o is very large compared to f
DAll lenses invert images regardless of object distance
Real images (s_i > 0) form when s_o > f for a converging lens. The magnification formula m = −s_i/s_o is negative whenever s_i and s_o have the same sign, which is always the case for real images. Negative m means inverted. Option D is wrong: when s_o < f, a converging lens produces a virtual, upright image (m > 0).
Question 3 True / False
A converging lens generally produces a real image.
TTrue
FFalse
Answer: False
A converging lens produces a virtual image whenever the object is placed inside the focal length (s_o < f). In that case, 1/s_i = 1/f − 1/s_o becomes negative, giving s_i < 0 — the image is on the same side as the object and cannot be projected on a screen. This is the magnifying glass mode. 'Converging lens' describes the lens geometry, not a guarantee of image type.
Question 4 True / False
A positive value of magnification (m > 0) from the formula m = −s_i/s_o generally means the image is larger than the object.
TTrue
FFalse
Answer: False
The sign of m encodes orientation, not size. Positive m means the image is upright (virtual); negative m means inverted (real). The magnitude |m| encodes size ratio. A virtual image could have m = +0.5 — upright and smaller than the object. Size and orientation are two independent pieces of information packed into a single signed number.
Question 5 Short Answer
Why does the thin lens equation predict that s_i approaches infinity as the object approaches the focal point, and what practical optical device exploits this behavior?
Think about your answer, then reveal below.
Model answer: As s_o → f, the term 1/f − 1/s_o → 0, so 1/s_i → 0 and s_i → ∞. Physically, rays from an object at the focal point exit the lens parallel — they never converge. A collimator or beam expander exploits this: placing a point source at the focal point produces a parallel output beam.
This limiting case is not a failure of the equation — it is a real physical result. Slide projectors, flashlights, and laser beam expanders all exploit it. The equation is continuous: as the object moves toward f from outside, s_i grows toward +∞ (real image pushed further away); as the object moves toward f from inside, s_i grows toward −∞ (virtual image pushed further away on the same side). The focal point is the singularity where the image 'goes to infinity.'