Ray diagrams provide useful information about the characteristics
Ray diagrams provide useful information about the characteristics of an image. They do not provide any quantitative information. eg: the distance and image size. We need to use the Mirror equation and the Magnification equation. Sample problem #1 10.0 cm tall candle is placed a distance of 14.0 cm from a
concave mirror having a focal length of 7.0cm Determine the image distance and the image size. 1/f = 1/do + 1/di 1/(7.0cm) = 1/(14.0 cm) + 1/di 14 1 - 1 = 1
di 7 14 di 2 1 1 = 14 1i
d 14 14 = 1 di = hi/ho = - di/do hi/10.0cm = 14.0cm/14.0cm hi = - 1 x 10.0cm
hi = - 10.0cm the image height is -10.0 cm. The negative value for image height indicate that the image is an inverted. Sample Problem #2 A 4.0-cm tall light bulb is placed a
distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image Given:size. ho = 4.0 cm do = 45.7 cm f = Unknowns: 15.2 cm di = ? hi = ?
1/f = 1/do + 1/di 1/(15.2 cm) = 1/(45.7 cm) + 1/di 0.0658 = 0.0219 + 1/ di 0.0439 = 1/di di = 22.8 cm the image distance is 22.8 cm. To determine the image height, the
magnification equation is needed hi/ho = - di/do hi /(4.0 cm) = - (22.8 cm)/(45.7 hi = - (4.0 cm) cm) * (22.8 cm)/(45.7 cm) hi = -1.99 cm the image height is -1.99 cm.
Sample Problem #3 The same light bulb is placed a distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image size. Given: ho = 4.0 cm
Unknowns: di = ? do = 8.3 cm cm hi = ? f = 15.2 1/f = 1/do + 1/di 1/(15.2 cm) = 1/(8.3 cm) + 1/di
0.0658 = 0.120 + 1/di -0.0547 = 1/di di = -18.3 cm the image distance is 18.3cm. The negative distance indicates that the image is virtual and located behind the mirrror. To determine the image height, the magnification equation is needed hi/ho = - di/do
hi /(4.0 cm) = - (-18.2 cm)/(8.3 hi = - (4.0cm) cm) * (-18.2 cm)/(8.3 cm) hi = 8.8 cm the image height is 8.8 cm. The positive value for image height indicate that the image is an upright or erect..
The sign conventions for the given quantities in the mirror and magnification equations f is +ve if the mirror is a concave mirror f is -ve if the mirror is a convex mirror di is + ve if the image is a real image and located on the object's side of the mirror.
d is - ve if the image is a virtual i image and located behind the mirror. hi is + ve if the image is an upright image hi is - ve if the image an inverted image
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