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WAEC Physics 2012 Paper

Question : 28

Total: 47

Solution:

Concept:

Object distance is less than focal length, so a converging lens forms a virtual, erect, and magnified image.

Explanation:

Given:

u = 10\,\text{cm}

and

f = 15\,\text{cm}

Since

u < f

, the image formed is virtual and on the same side as the object.

Use the lens formula:

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

Substitute the values:

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

Therefore,

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

So,

v = -30\,\text{cm}

. The negative sign shows the image is virtual.

Magnification,

m = \frac{|v|}{u} = \frac{30}{10} = 3.0

Answer:

3.0

(Option A).

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