A person’s face is 30.0 cm in front of a concave makeup mirror, producing an upright image that is 1.45 times as large as the object. What is the mirror’s focal length f?
--answer in cm
Before anyone says do your own homework, I know that the object distance = do = 30.0cm and the
image height= 1.44 times *h0
The magnification equation is:
m = - di/do =hi/ ho
The mirror equation is:
1/d0 + 1/di = 1/f
I keep getting the wrong answer- plz help, and it would be nice if you exlpained it also- thanks
--answer in cm
Before anyone says do your own homework, I know that the object distance = do = 30.0cm and the
image height= 1.44 times *h0
The magnification equation is:
m = - di/do =hi/ ho
The mirror equation is:
1/d0 + 1/di = 1/f
I keep getting the wrong answer- plz help, and it would be nice if you exlpained it also- thanks
-
As the image is upright and magnified it is virtual (formed behind the mirror) so image distance di is negative.(-di)
Magnification .. m = 1.45 = di/do** .. .. di = 1.45 x 30cm = (-)43.5 cm
Using .. 1/do + 1/-di = 1/f
1/30 - 1/43.5 = 1/f
1/f = 0.01034 .. .. .. ►f = 96.70 cm
Magnification .. m = 1.45 = di/do** .. .. di = 1.45 x 30cm = (-)43.5 cm
Using .. 1/do + 1/-di = 1/f
1/30 - 1/43.5 = 1/f
1/f = 0.01034 .. .. .. ►f = 96.70 cm