• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Problems with negatives.

PrincePhoenix

Gold Member
116
2
1st problem:
A small object of size 0.5 mm is placed close to a convex lens of focal length 5cm. The virtual image formed is at a distance of 25 cm from the lens. Find magnification.
Data:
focal length, f = 5cm
distance of image , q = -25 cm (because image is virtual)
distance of object , p = ?
magnification, M = ?




1/f = 1/p + 1/q
M = height of image/height of object = q/p
(ratio between height and distance of image and object)





Solution:
First finding p,
1/f = 1/p + 1/q
1/p = 1/f - 1/q
1/p = 1/5 - (1/-25)
..................
1/p = 6/25
p = 25/6 = 4.16cm
Now putting value of 'p' in magnification formula,
M = height of image/height of object = q/p
(ratio between height and distance of image and object)

M = -25/4.16
M = -6
Can magnification be -ve? It is so in this problem. What have I done wrong?


2nd problem:
1-This part of a numerical problem as our teacher did it in classroom.
Data:
distance of object, p = 20cm
focal length of concave lens,f = -19 (Is focal length of concave lens really always -ve?)
distance of image, q = ?
Solution:
According to lens formula,
1/f = 1/p + 1/q
=> 1/q = 1/f - 1/p
1/q = 1/-19 - 1/20
1/q = -20-19/380
1/q = -39/380
q = -9.74cm
Why can't LCM be negative at line 4,5 of solution when one of the denominators is -ve? And is the -ve value of distance of image correct?
 

Redbelly98

Staff Emeritus
Science Advisor
Homework Helper
12,037
125
1st problem:
A small object of size 0.5 mm is placed close to a convex lens of focal length 5cm. The virtual image formed is at a distance of 25 cm from the lens. Find magnification.
Data:
focal length, f = 5cm
distance of image , q = -25 cm (because image is virtual)
distance of object , p = ?
magnification, M = ?




1/f = 1/p + 1/q
M = height of image/height of object = q/p
(ratio between height and distance of image and object)
It should be -q/p in that equation.

Solution:
First finding p,
1/f = 1/p + 1/q
1/p = 1/f - 1/q
1/p = 1/5 - (1/-25)
..................
1/p = 6/25
p = 25/6 = 4.16cm
Now putting value of 'p' in magnification formula,
M = height of image/height of object = q/p
(ratio between height and distance of image and object)

M = -25/4.16
M = -6
Can magnification be -ve? It is so in this problem. What have I done wrong?
See previous note.
By the way, magnification can be negative, and is when q and p are both positive.

2nd problem:
1-This part of a numerical problem as our teacher did it in classroom.
Data:
distance of object, p = 20cm
focal length of concave lens,f = -19 (Is focal length of concave lens really always -ve?)
distance of image, q = ?
Solution:
According to lens formula,
1/f = 1/p + 1/q
=> 1/q = 1/f - 1/p
1/q = 1/-19 - 1/20
1/q = -20-19/380
1/q = -39/380
q = -9.74cm
Why can't LCM be negative at line 4,5 of solution when one of the denominators is -ve? And is the -ve value of distance of image correct?
The LCM could be -380, but most people consider it easier to move the negative sign to the numerator and use a positive denominator.

For a negative denominator, this becomes

1/(-19) - 1/20
= 20/(-380) + 19/(-380)
= (+20 +19)/(-380)
= ___?
 

PrincePhoenix

Gold Member
116
2
Thanks.
 

PrincePhoenix

Gold Member
116
2
What does negative magnification mean anyway?
 

Redbelly98

Staff Emeritus
Science Advisor
Homework Helper
12,037
125
It means the image is inverted, relative to the object. (Their heights have opposite signs.)
 

Related Threads for: Problems with negatives.

Replies
4
Views
2K
Replies
6
Views
12K
Replies
15
Views
620
  • Posted
Replies
7
Views
467

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top