A. Focal length = half of radius of curvature = > f = 25/2=12.5 cm
B. 1/o+1/i=1/f, where o is the object distance, I is the image distance , and f is the focal length,
1/I =1/f-1/o= 1/12.5- 1/40 = > I = 18.2 cm.
C. Magnification = 12.5/(40-12.5) = 0.45
D. Height of image is 10•0.45= 4.5 cm
E. Object is outside the focal lenth => real and invert
The distance between a 10 cm tall candle and the reflecting surface of a concave mirror is 40 cm. The radius of curvature of the mirror is 25 cm. Now,
A. What is the focal length of the mirror?
B. What is the image distance?
C. What is the magnification of the image?
D. What is the height of the image?
E. State whether the image is real or virtual and whether it is upright or inverted.
I'm honestly so confused
2 answers
A. Focal length = half of radius of curvature = > f = 25/2=12.5 cm
B. 1/di + 1/do = 1/f
1/di + 1/-40 = 1/-12.5
di = 18.18m
C. M = -di/do = 18.18/-40 = -.45
Since the magnitude of image M is less than 1, the image is diminished.
D. M = hi/ho
hi = M*ho = -.45 * 10= -4.5
E. The negative sign of image distance indicates that the image is formed in front of the mirror and so is real. The negative sign of height of image indicates that the image is inverted.
B. 1/di + 1/do = 1/f
1/di + 1/-40 = 1/-12.5
di = 18.18m
C. M = -di/do = 18.18/-40 = -.45
Since the magnitude of image M is less than 1, the image is diminished.
D. M = hi/ho
hi = M*ho = -.45 * 10= -4.5
E. The negative sign of image distance indicates that the image is formed in front of the mirror and so is real. The negative sign of height of image indicates that the image is inverted.