An object placed in front of a convex lens, forms an image of same size on a screen. Moving the object 12 cm closer to the lens results in the formation of a real image which is three times the size of the object. Calculate the focal length of the lens.

Let the focal length be f.

FIRST POSITION — the image is the same size as the object. A convex lens forms a real, same-sized image only when the object is at 2F, so u₁ = −2f

SECOND POSITION — the object moves 12 cm closer, and the real image is three times the size. u₂ = −(2f − 12) For a real, inverted image the magnification is negative, so m = −3: m = v₂ ÷ u₂ = −3 v₂ = −3 u₂ = −3 × −(2f − 12) = 3(2f − 12)

Substitute into the lens formula 1/v − 1/u = 1/f: 1 ÷ [3(2f − 12)] − 1 ÷ [−(2f − 12)] = 1/f 1 ÷ [3(2f − 12)] + 1 ÷ (2f − 12) = 1/f

Taking a common denominator of 3(2f − 12): (1 + 3) ÷ [3(2f − 12)] = 1/f 4 ÷ [3(2f − 12)] = 1/f

Cross-multiplying: 4f = 3(2f − 12) 4f = 6f − 36 36 = 6f − 4f 36 = 2f f = 18 cm

Focal length of the lens = 18 cm

The sentence that unlocks this is "an image of the same size on a screen". On a SCREEN means the image is real, and a convex lens gives a real same-sized image at exactly one place: the object at 2F and the image at 2F. So the first object distance is 2f, and that is what lets you write the second one as 2f − 12.

The magnification is −3, not +3, because the image is real and therefore inverted. Using +3 gives the wrong sign chain and a wrong focal length.

Check the answer by going back: f = 18, so the object starts at 36 cm and moves to 24 cm. At u = −24, the lens formula gives 1/v = 1/18 − 1/24 = (4 − 3)/72 = 1/72, so v = 72 cm, and m = 72 ÷ (−24) = −3 ✓. Three times the size, real and inverted, exactly as stated.