An object is placed 12\text{ cm} in front of a diverging lens having a focal length of 18\… — JAMB UTME

An object is placed $12\text{ cm}$ in front of a diverging lens having a focal length of $18\text{ cm}$ . What is the magnitude of the image distance from the lens?

$7.2\text{ cm}$ Answer
B
$36.0\text{ cm}$
C
$30.0\text{ cm}$
D
$6.0\text{ cm}$

Answer

The magnitude of the image distance is

7.2\text{ cm}

By the thin lens equation

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

, substituting

f = -18\text{ cm}

(due to the lens being diverging) and

u = +12\text{ cm}

gives

\frac{1}{v} = -\frac{1}{18} - \frac{1}{12} = -\frac{5}{36}\text{ cm}^{-1}

, which yields

v = -7.2\text{ cm}

. Thus, the image is located

7.2\text{ cm}

from the lens.

Step-by-Step Solution

Identify the given values and assign the correct signs according to the real-is-positive sign convention.

Object distance

u = +12\text{ cm}

; Focal length of diverging (concave) lens

f = -18\text{ cm}

Diverging lenses have virtual focus, so their focal length must be taken as negative.

Set up the thin lens formula

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

to solve for image distance

v

\frac{1}{-18} = \frac{1}{12} + \frac{1}{v} \implies \frac{1}{v} = -\frac{1}{18} - \frac{1}{12}

Rearranging the equation isolates the reciprocal of the image distance.

Find a common denominator and solve for

v

\frac{1}{v} = \frac{-2 - 3}{36} = -\frac{5}{36}\text{ cm}^{-1} \implies v = -\frac{36}{5} = -7.2\text{ cm}

The negative sign indicates that the image formed is virtual and located on the same side of the lens as the object.

Take the magnitude of the image distance.

|v| = 7.2\text{ cm}

The question asks for the distance, which is a scalar magnitude.

Key Concept

Thin Lens Formula & Sign Convention for Concave Lenses

Estimated Time:1m 30s

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