An object is placed 10 cm from a concave mirror; the image is virtual and three

Question

An object is placed 10 cm from a concave mirror; the image is virtual and three times the size of the object. What is the focal length of the mirror? 7.5 cm 15 cm 20 cm 30 cm When an object is placed 10 cm (0.10m) from a converging lens, the real image formed is twice the size of the object. What is the power of this lens? 6.7 D 10 D 15 D 20 D Two converging lens are placed in contact. If one has a focal length of 10 cm, and the other has a focal length of 20 cm, what is the power of the combination? 10 D 15 D 20 D 30 D We view an object at various distances using a mirror with focal length 12 m. If the object is 6 m in front of the mirror, where is the image? 6 m in front of the mirror. 6 m behind the mirror. 12 m in front of the mirror. 12 m behind the mirror. If the object is 6 m in front of the mirror, what is its magnification? It is half as large, and the image is upright. It is half as large, and the image is inverted. It is twice as large, and the image is upright. It is twice as large, and the image is inverted. If the object is 24 m away from the mirror, where is the image? 12 m in front of the mirror. 12 m behind the mirror. 24 m in front of the mirror. 24 m behind the mirror.

Explanation / Answer

Problem 7)

Magnification M = hi/ho , where hi is the height of the image , and ho is the height of the object.

hi = 3 ho, because the image is three times the size of the object.

M = 3 ho/ho = 3

The The Gaussian mirror equation is

(1/o) + (1/i) = 1/f,    where o is the object distance = 10 cm, i is the image distance, and f is the focal length

But also M = - i/o

3 = - i/10

i = - 30 cm (virtual image, M is positive for upright image

Substitute in the mirror equation:

(1/10) - (1/30) = 1/f

so f = 15 cm [Choice b]

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