Show Two Positions of Converging Lens for Sharp Image Formation

[leolaw]

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 d_T = d_i + d_0 between the object and the screen is kept fixed, but the lens can be moved.

Show that if d_t > 4f , there will be two positions where the lens can be placed and a sharp image can be produced on the screen.
And if d_t < 4f, 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 don't seem to know how to tackle this problem

 

[GCT]

You'll need to do some rearranging of equations and plugging in, try finding dt as a function of f.

 

[Andrew Mason]

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 d_T = d_i + d_0 between the object and the screen is kept fixed, but the lens can be moved.

Show that if d_t > 4f , there will be two positions where the lens can be placed and a sharp image can be produced on the screen.
And if d_t < 4f, 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 don't seem to know how to tackle this problem

From the lens equation:

\frac{1}{f} = \frac{1}{i} + \frac{1}{o}

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

\frac{1}{f} = \frac{1}{i} + \frac{1}{S-i}

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

i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}

AM

 

[leolaw]

after \frac{1}{f} = \frac{1}{i} + \frac{1}{S-i} , I have f = \frac{si - i^2}{s} , and then i set the equation equals to zero: i^2 + fs - si = 0, but i don't get how you can solve for i from i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}.

 

[Andrew Mason]

after \frac{1}{f} = \frac{1}{i} + \frac{1}{S-i} , I have f = \frac{si - i^2}{s} , and then i set the equation equals to zero: i^2 + fs - si = 0, but i don't get how you can solve for i from i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}.

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

 

[leolaw]

for i to be real and have two solutions, \sqrt{S^2 - 4Sf} must be greater than 0.
So, s^2 - 4sf > 0

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

But I don't know how you can get i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}, from i^2 + fs - si = 0

 

[Andrew Mason]

for i to be real and have two solutions, \sqrt{S^2 - 4Sf} must be greater than 0.
So, s^2 - 4sf > 0

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

But I don't know how you can get i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}, from i^2 + fs - si = 0

That is just the quadratic formula. The general quadratic equation:

ax^2 + bx + c = 0

has solutions:

x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

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

AM

 

[leolaw]

That is just the quadratic formula. The general quadratic equation:

ax^2 + bx + c = 0

has solutions:

x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

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!

 

[Andrew Mason]

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