A thin diverging lens has a focal length of 20\text{ cm}. If it forms an upright image tha… — JAMB UTME

A thin diverging lens has a focal length of $20\text{ cm}$ . If it forms an upright image that is one-fourth the size of the object, what is the distance of the object from the lens?

A
$25\text{ cm}$
$60\text{ cm}$ Answer
C
$100\text{ cm}$
D
$15\text{ cm}$

Answer

The distance of the object from the lens is

60\text{ cm}

For a diverging (concave) lens, the focal length is negative (

f = -20\text{ cm}

). Since a diverging lens always forms a virtual and upright image, the image distance

v

is negative. Given that linear magnification

m = \frac{|v|}{u} = \frac{1}{4}

, we obtain

v = -\frac{u}{4}

. Substituting these values into the lens formula

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

gives

\frac{1}{-20} = \frac{1}{u} - \frac{4}{u} = -\frac{3}{u}

, which solves to

u = 60\text{ cm}

Step-by-Step Solution

Identify given parameters and apply optical sign conventions.

Focal length of diverging lens

f = -20\text{ cm}

. Linear magnification

m = \frac{1}{4}

Diverging (concave) lenses always have a negative focal length.

Express image distance

v

in terms of object distance

u

v = -\frac{u}{4}

Magnification

m = \frac{|v|}{u} = \frac{1}{4}

, and since diverging lenses produce virtual images,

v

must be negative.

Substitute

f = -20\text{ cm}

and

v = -\frac{u}{4}

into the thin lens formula

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

\frac{1}{-20} = \frac{1}{u} + \frac{1}{-\frac{u}{4}} = \frac{1}{u} - \frac{4}{u} = -\frac{3}{u}

Combining terms under a common denominator

u

simplifies the expression.

Solve for the object distance

u

-\frac{1}{20} = -\frac{3}{u} \implies u = 3 \times 20 = 60\text{ cm}

Cross-multiplying yields the positive object distance of

60\text{ cm}

Key Concept

Thin Lens Formula and Optical Sign Conventions for Diverging Lenses

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