solve at least three numericals of mirror concave lens
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Answer:
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Answer:
An object is placed 10 cm from a concave mirror. The focal length is 5 cm. Determine (a) The image distance (b) the magnification of image
Known :
The focal length (f) = 5 cm
The object distance (do) = 10 cm
Solution :
Formation of image by concave mirror :
The image distance :
1/di = 1/f – 1/do = 1/5 – 1/10 = 2/10 – 1/10 = 1/10
di = 10/1 = 10 cm
The image distance is 10 cm.
The magnification :
m = –di / do = -10/10 = -1
1 means that the image is the same as the object.
The minus sign indicates that the image is inverted. If the sign is positive than the image is upright.
2.An image an by a concave mirror is 4 times greater than the object. If the radius of curvature 20 cm, determine the object distance in front of the mirror!
Known :
The magnification of image (m) = 4
The radius of curvature (r) = 20 cm
The focal length (f) = r/2 = 20/2 = 10 cm
Wanted : The object distance (do)
Solution :
m = – di / do
4 = – di / do
– di = 4 do
di = – 4 do
1/f = 1/do + 1/di
1/10 = 1/do + 1/4do
4/40 = 4/4do + 1/4do
4/40 = 5/4do
(4)(4s) = (5)(40)
16 do = 200
do = 12.5 cm
The object distance = 12.5 cm.
3.A 1-cm high object is placed 10 cm from a concave mirror with the focal length, f = 15 cm. Determine :
A. The image distance ?
B. The image height?
C. The properties of image formed by the concave mirror?
Known :
The object height (h) = 1 cm
The object distance (do) = 10 cm
The focal length of the concave mirror (f) = 15 cm
Solution :
A. The image distance (di)
1/f = 1/do + 1/di
1/di = 1/f – 1/do = 1/15 – 1/10 = 2/30 – 3/30 = -1/30
di = -30/1 = -30 cm
The negative sign indicates that the image is virtual or the image is behind the mirror.
B. The image height (hi)
The magnification of the image (M) :
M = -di/do = hi/ho
M = -(-30)/10 = 30/10 = 3 times
The image height (hi) :
M = hi / ho
3 = hi / 1 cm
hi = 3 (1 cm)
hi = 3 cm
the image height is 3 cm. The plus sign indicates that the image upward.
C. The properties of the image :
Virtual, upward, larger than object
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