A thin converging lens of focal length 15\text{ cm} forms an erect image that is magnified… — JAMB UTME

A thin converging lens of focal length $15\text{ cm}$ forms an erect image that is magnified three times. What is the distance of the object from the lens?

$10\text{ cm}$ Answer
B
$20\text{ cm}$
C
$5\text{ cm}$
D
$45\text{ cm}$

Answer

10\text{ cm}

An erect image produced by a thin converging lens is virtual, which means the linear magnification is positive (

m = +3

) and the image distance is negative relative to the real object (

v = -3u

). Substituting

f = +15\text{ cm}

and

v = -3u

into the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

gives

\frac{1}{15} = \frac{1}{u} - \frac{1}{3u} = \frac{2}{3u}

. Solving for

u

yields

u = 10\text{ cm}

Step-by-Step Solution

Determine the nature of the image and establish the relationship between image distance and object distance

Since the image formed by a converging lens is erect, it must be virtual. Thus, linear magnification

m = +3 = -\frac{v}{u}

, giving

v = -3u

A single convex lens produces an erect image only when the image is virtual, requiring a negative image distance under standard optical sign conventions.

Substitute focal length and image distance into the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{15} = \frac{1}{u} + \frac{1}{-3u}

The thin lens equation relates focal length, object distance, and image distance.

Simplify the algebraic expression and solve for object distance u

\frac{1}{15} = \frac{3 - 1}{3u} = \frac{2}{3u} \implies 3u = 30 \implies u = 10\text{ cm}

Finding a common denominator allows direct solution for the object distance

u

Key Concept

Thin lens formula and sign conventions for virtual images

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