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