A concave mirror forms an image on a screen that is twice as large as the object. Both the object and the screen are then moved to produce an image on the screen that is three times the size of the object. If the screen is moved 75 cm in the process, how far is the object moved? What is the focal length of the mirror? Solution magnification: m = hi / ho = di/do hi = size of the image ho = size of the object but, hi = 2* ho, so m = 2*ho / h o = 2, Then di= 2*do mirror formula: 1/ f = 1/do + 1/di 1/f = 1/do + 1/2do 1/f = 3/2do new magnification is m \' = 3, so distance moved by the screen is s = 75 cm new image distance: di \' = di +s do\' =do-s\' then, m \' = di\'/do\' = di +s /do-s\', therefore, 3(do-s\') = di +s new focal length: 1/ f = 1/do \' + 1/di\' 1/f = 1/ do-s\' + 1/3(do-s\') solve the above equations, we get, s\' = do /9 thus, do =112.5 cm, hence, s\'= 112.5cm /9 =12.5 cm therefore, the focal length: f = 3/2do = 75 cm.

1. A concave mirror forms an image on a screen that is twice as large as the object. Both the object
and the screen are then moved to produce an image on the screen that is three times the size of
the object. If the screen is moved 75 cm in the process, how far is the object moved? What is the
focal length of the mirror?
Solution
magnification: m = hi / ho = di/do
hi = size of the image
ho = size of the object
but, hi = 2* ho, so m = 2*ho / h o = 2, Then di= 2*do
mirror formula:
1/ f = 1/do + 1/di
1/f = 1/do + 1/2do
1/f = 3/2do
new magnification is m ' = 3, so distance moved by the screen is s = 75 cm
new image distance:
di ' = di +s
do' =do-s'
then, m ' = di'/do' = di +s /do-s', therefore, 3(do-s') = di +s
new focal length:
1/ f = 1/do ' + 1/di'
1/f = 1/ do-s' + 1/3(do-s')
solve the above equations, we get, s' = do /9
thus, do =112.5 cm, hence, s'= 112.5cm /9 =12.5 cm
therefore, the focal length:
f = 3/2do = 75 cm