An object is placed 10\text{ cm} in front of a converging lens of focal length 15\text{ cm… — JAMB UTME

An object is placed $10\text{ cm}$ in front of a converging lens of focal length $15\text{ cm}$ . What is the distance of the image from the lens and its nature?

$30\text{ cm}$ from the lens and virtualAnswer
B
$30\text{ cm}$ from the lens and real
C
$6\text{ cm}$ from the lens and real
D
$6\text{ cm}$ from the lens and virtual

Answer

The image is

30\text{ cm}

from the lens and virtual.

Using the thin lens equation

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

with

f = +15\text{ cm}

and

u = +10\text{ cm}

yields

\frac{1}{v} = \frac{1}{15} - \frac{1}{10} = -\frac{1}{30}\text{ cm}^{-1}

. Taking the reciprocal gives

v = -30\text{ cm}

. The magnitude of the image distance is

30\text{ cm}

, and the negative sign confirms that the image formed is virtual.

Step-by-Step Solution

Identify the given optical parameters and sign conventions.

Object distance

u = +10\text{ cm}

, Focal length for converging lens

f = +15\text{ cm}

By convention for real objects and converging lenses, both

u

and

f

are positive.

Substitute the values into the thin lens formula

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

\frac{1}{15} = \frac{1}{10} + \frac{1}{v}

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

Rearrange to solve for the image distance reciprocal

\frac{1}{v}

\frac{1}{v} = \frac{1}{15} - \frac{1}{10} = \frac{2 - 3}{30} = -\frac{1}{30}\text{ cm}^{-1}

Subtracting

\frac{1}{10}

from

\frac{1}{15}

requires finding a common denominator of 30.

Take the reciprocal to find

v

and interpret its sign.

v = -30\text{ cm}

The magnitude of the distance is

30\text{ cm}

. The negative sign indicates that the image is formed on the same side as the object, making it virtual.

Key Concept

Thin lens formula and sign convention for image formation

Estimated Time:1m 30s

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