Question
a converging lens has a focal length of 15cm. Find two positions where a real object can produce images four times its size.
Answers
drwls
Di = 4 Do, or
Di = -4 Do
will both work.
Di = image distance
Do = object distance
First case:
1/Di + 1/Do = 1/f = 1/15
1/(4Do) + 1/Do = 1/15
5/(4Do) = 1/15
Do = 75/4 = 18.75 cm
Second case:
-1/4Do + 1/Do = 1/15
3/(4Do) = 1/15
4 Do = 45
Do = 11.25 cm
Di = -4 Do
will both work.
Di = image distance
Do = object distance
First case:
1/Di + 1/Do = 1/f = 1/15
1/(4Do) + 1/Do = 1/15
5/(4Do) = 1/15
Do = 75/4 = 18.75 cm
Second case:
-1/4Do + 1/Do = 1/15
3/(4Do) = 1/15
4 Do = 45
Do = 11.25 cm