Converging lens

  • Thread starter leolaw
  • Start date
  • #1
85
1
A bright object is placed on one sid eof a converging lens of focal length f, and a white screen for viewing the image is on the opposite side. The distance [tex]d_T = d_i + d_0 [/tex] between the object and the screen is kept fixed, but the lens can be moved.

Show that if [tex]d_t > 4f[/tex] , there will be two positions where the lens can be placed and a sharp image can be produced on the screen.
And if [tex]d_t < 4f [/tex], no lens position where a shakrp image is formed.
Also determine a formula for the distance b/w the two lens position,and the ratio of the image sizes.

the attachement is the work that i was trying figure out how to do it. but i dont seem to know how to tackle this problem
 

Attachments

Answers and Replies

  • #2
GCT
Science Advisor
Homework Helper
1,728
0
You'll need to do some rearranging of equations and plugging in, try finding dt as a function of f.
 
  • #3
Andrew Mason
Science Advisor
Homework Helper
7,652
383
leolaw said:
A bright object is placed on one sid eof a converging lens of focal length f, and a white screen for viewing the image is on the opposite side. The distance [tex]d_T = d_i + d_0 [/tex] between the object and the screen is kept fixed, but the lens can be moved.

Show that if [tex]d_t > 4f[/tex] , there will be two positions where the lens can be placed and a sharp image can be produced on the screen.
And if [tex]d_t < 4f [/tex], no lens position where a shakrp image is formed.
Also determine a formula for the distance b/w the two lens position,and the ratio of the image sizes.

the attachement is the work that i was trying figure out how to do it. but i dont seem to know how to tackle this problem
From the lens equation:

[tex]\frac{1}{f} = \frac{1}{i} + \frac{1}{o} [/tex]

since S = i + o (S = object to screen distance):

[tex]\frac{1}{f} = \frac{1}{i} + \frac{1}{S-i} [/tex]

This gives you a quadratic equation in terms of i. Solve that using the quadratic formula and you should get two solutions for i:

[tex]i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}[/tex]

AM
 
Last edited:
  • #4
85
1
after [tex]\frac{1}{f} = \frac{1}{i} + \frac{1}{S-i} [/tex] , I have [tex] f = \frac{si - i^2}{s} [/tex] , and then i set the equation equals to zero: [tex] i^2 + fs - si = 0 [/tex], but i dont get how you can solve for i from [tex]i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}[/tex].
 
  • #5
Andrew Mason
Science Advisor
Homework Helper
7,652
383
leolaw said:
after [tex]\frac{1}{f} = \frac{1}{i} + \frac{1}{S-i} [/tex] , I have [tex] f = \frac{si - i^2}{s} [/tex] , and then i set the equation equals to zero: [tex] i^2 + fs - si = 0 [/tex], but i dont get how you can solve for i from [tex]i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}[/tex].
Well, you can't solve for i unless you know S. But that is not what the question asks.

What is the condition for i to be real? What is the condition for i to have 2 real values?

AM
 
  • #6
85
1
for i to be real and have two solutions, [tex]\sqrt{S^2 - 4Sf}[/tex] must be greater than 0.
So, [tex] s^2 - 4sf > 0 [/tex]

[tex] s^2 > 4sf [/tex]
[tex] s > 4f [/tex] and we set [tex] s = i + o [/tex] before, so we get the answer for the first two questiosn.

But I don't know how you can get [tex]i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}[/tex], from [tex] i^2 + fs - si = 0 [/tex]
 
  • #7
Andrew Mason
Science Advisor
Homework Helper
7,652
383
leolaw said:
for i to be real and have two solutions, [tex]\sqrt{S^2 - 4Sf}[/tex] must be greater than 0.
So, [tex] s^2 - 4sf > 0 [/tex]

[tex] s^2 > 4sf [/tex]
[tex] s > 4f [/tex] and we set [tex] s = i + o [/tex] before, so we get the answer for the first two questiosn.

But I don't know how you can get [tex]i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}[/tex], from [tex] i^2 + fs - si = 0 [/tex]
That is just the quadratic formula. The general quadratic equation:

[tex]ax^2 + bx + c = 0[/tex]

has solutions:

[tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

In this case, a = 1, b = -S, c = sf

AM
 
  • #8
85
1
Andrew Mason said:
That is just the quadratic formula. The general quadratic equation:

[tex]ax^2 + bx + c = 0[/tex]

has solutions:

[tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

In this case, a = 1, b = -S, c = sf

AM
AHAHHAHHAHAHA. I feel myself really stupid now....can't even find the coeiffient term of a quadratic equation!
 
  • #9
Andrew Mason
Science Advisor
Homework Helper
7,652
383
leolaw said:
AHAHHAHHAHAHA. I feel myself really stupid now....can't even find the coeiffient term of a quadratic equation!
That happens during exam time! Ease up .. it was a bit of a tricky question.

AM
 

Related Threads on Converging lens

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
879
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
8
Views
2K
Replies
4
Views
6K
Replies
3
Views
3K
  • Last Post
Replies
1
Views
2K
Top