Questions: The Thin Lens Equation

Question 1

An object is placed 10 cm from a converging lens with a focal length of 15 cm. Where does the image form?

• A. 30 cm on the far (transmission) side of the lens — a real, inverted image
• B. 30 cm on the same side as the object — a virtual, upright image
• C. 6 cm on the far side of the lens — a real, inverted image
• D. No image forms because the object is inside the focal length

Answer: (B)

Using 1/dᵢ = 1/f − 1/dₒ = 1/15 − 1/10 = 2/30 − 3/30 = −1/30, so dᵢ = −30 cm. The negative sign means the image is on the same side as the object — a virtual image. This is the magnifying glass configuration: when the object is closer than the focal length (dₒ < f), dᵢ goes negative, producing an upright, magnified virtual image. Option D is wrong because an image does form — it's just virtual.

Question 2

What happens to the image distance dᵢ as an object is moved from far away toward the focal point of a converging lens (approaching from dₒ > f)?

• A. dᵢ decreases toward zero as the image gets closer to the lens
• B. dᵢ remains constant at twice the focal length
• C. dᵢ increases without bound — the image recedes toward infinity
• D. dᵢ becomes negative, indicating the image flips to virtual before dₒ reaches f

Answer: (C)

As dₒ → f from above, 1/dₒ → 1/f, so 1/dᵢ = 1/f − 1/dₒ → 0, meaning dᵢ → ∞. The image races away from the lens indefinitely. At exactly dₒ = f, the rays exit parallel and no image forms at a finite distance. Option D describes what happens when dₒ passes through f and becomes less than f — the image then flips to virtual — but that is not what happens as dₒ approaches f from above.

Question 3

A diverging lens can form a real image if the object is placed far enough from the lens.

• T. True
• F. False

Answer: False

For a diverging lens, f < 0. Substituting into 1/dᵢ = 1/f − 1/dₒ: since both 1/f and −1/dₒ are negative (for any positive dₒ), dᵢ is always negative. A diverging lens always produces a virtual image — upright, reduced, and on the same side as the object — regardless of object distance. There is no object placement that produces a real image through a single diverging lens.

Question 4

The lateral magnification m = −dᵢ/dₒ predicts that real images formed by a single converging lens are always inverted.

• T. True
• F. False

Answer: True

For a real image formed by a converging lens, the image is on the far (transmission) side, so dᵢ > 0. Object distance dₒ is also positive. Therefore m = −dᵢ/dₒ is negative, indicating an inverted image. This is always true for real images through a single lens: real images are inverted. Virtual images (dᵢ < 0) have positive m and are upright — this is why a magnifying glass and the human eye both produce upright virtual images when used normally.

Question 5

A student holds a converging lens close to a page of text and sees an upright, enlarged image of the letters. Explain using the thin lens equation why this image must be virtual.

Model answer: For the image to be upright and magnified through a single converging lens, the object must be inside the focal length (dₒ < f). Substituting into 1/dᵢ = 1/f − 1/dₒ: when dₒ < f, 1/dₒ > 1/f, so 1/dᵢ < 0, meaning dᵢ < 0. A negative image distance means the image forms on the same side as the object — not where light actually converges, but where diverging rays appear to come from when you look through the lens. That is the definition of a virtual image.

The sign of dᵢ is the key: positive means a real image (rays actually converge on the far side), negative means a virtual image (rays diverge, and the image is where they appear to originate). For a magnifying glass, the lens is too close to the object for rays to converge on the far side — instead they exit the lens still diverging, and your eye traces them back to an apparent source that is upright and enlarged. The thin lens equation automatically captures this: whenever dₒ < f for a converging lens, dᵢ turns negative.