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WAEC SSCE - Physics - 1990 (Objective)

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A concave mirror of radius of curvature 20cm has a pin placed at 15cm from its pole. What will be the magnification of the image formed?
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When an object is placed in front of a concave mirror, the mirror forms an image of the object. The image can either be real or virtual, and it can be magnified or diminished depending on the location of the object and the focal length of the mirror. In this case, the concave mirror has a radius of curvature of 20cm, which means its focal length is half of that or 10cm. The pin is placed at a distance of 15cm from the pole of the mirror, which means it is located beyond the focal point. Since the object is placed beyond the focal point, the image formed will be real and inverted. The magnification of the image can be calculated using the formula: magnification = height of image / height of object = -v/u where v is the distance of the image from the mirror, u is the distance of the object from the mirror, and the negative sign indicates that the image is inverted. Using the mirror formula (1/f = 1/v + 1/u), we can find the distance of the image from the mirror: 1/f = 1/v + 1/u 1/10 = 1/v + 1/15 v = 30cm Now, we can use the magnification formula to find the magnification of the image: magnification = height of image / height of object = -v/u magnification = -30/15 magnification = -2.00 The negative sign indicates that the image is inverted, and the magnification is 2.00. Therefore, the answer is 2.00.
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