# Show Two Positions of Converging Lens for Sharp Image Formation

• leolaw
In summary: Haha thanks for your help anyways! You're welcome! I'm glad I could help you figure it out. Just remember to always check your equations and make sure you know what the question is asking before jumping to solve. Good luck on your exam!In summary, the conversation discusses the use of a converging lens and a white screen for viewing images. It is shown that there are two positions where the lens can be placed to produce a sharp image on the screen when the distance between the object and the screen is greater than 4 times the focal length of the lens. If the distance is less than 4 times the focal length, no lens position will produce a sharp image. The formula for the distance between the two lens positions
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

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

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 $$d_T = d_i + d_0$$ 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

Last edited:
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 $$i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}$$.

leolaw said:
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

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 $$i^2 + fs - si = 0$$

leolaw said:
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 $$i^2 + fs - si = 0$$

$$ax^2 + bx + c = 0$$

has solutions:

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

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

AM

Andrew Mason said:

$$ax^2 + bx + c = 0$$

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!

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

## 1. What is a converging lens?

A converging lens is a type of lens that is thicker in the middle and thinner at the edges. It is also known as a convex lens and is commonly used in magnifying glasses, telescopes, and cameras.

## 2. How does a converging lens form a sharp image?

A converging lens works by refracting light rays that pass through it. When parallel light rays pass through a converging lens, they bend towards the center of the lens, eventually meeting at a point called the focal point. This is where a sharp image is formed.

## 3. What are the two positions of a converging lens for sharp image formation?

The two positions of a converging lens for sharp image formation are the object position and the image position. The object position is where the object is placed in front of the lens, and the image position is where the image is formed behind the lens.

## 4. How can I determine the object and image positions for a converging lens?

The object and image positions for a converging lens can be determined using the thin lens equation: 1/f = 1/do + 1/di, where f is the focal length of the lens, do is the object distance, and di is the image distance. This equation can also be used to calculate the magnification of the image.

## 5. What factors affect the sharpness of an image formed by a converging lens?

The sharpness of an image formed by a converging lens can be affected by several factors, including the distance between the object and the lens, the distance between the lens and the image, and the curvature of the lens. Additionally, the type of light source and the quality of the lens can also impact the sharpness of the image.

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