Problems with negatives.

  • #1
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?
 

Answers and Replies

  • #2
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,117
154
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)
= ___?
 
  • #4
PrincePhoenix
Gold Member
116
2
What does negative magnification mean anyway?
 
  • #5
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,117
154
It means the image is inverted, relative to the object. (Their heights have opposite signs.)
 

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