Solving Lens Question Homework: Converging, -0.582 m, Real Image

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In summary, converging in a lens question refers to light rays being brought closer together to form an image, while -0.582 m represents the focal length of the lens. To determine if an image formed by a lens is real or virtual, you must check if the light rays actually converge or only appear to converge. To solve a lens question with given values, you must identify the given values, determine the type of lens, use the thin lens equation to find the unknown value, and check your answer for accuracy.
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Hyari
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Homework Statement


"A certain lens focuses an object 2.85 m away as an image 48.3 cm on the other side of the lens. What type of lens is it and what is its focal length? Is the image real or virtual?


Homework Equations


1/f = 1/do + 1/di



The Attempt at a Solution


I'm guessing it's a converging lens since the image is on the other side. Diverging lenses create images on the side that light is coming from? Correct?

Attempt at finding focal length:

Since the image is coming from where light is not, that's a -1/di, the object is from where the light is coming from so it's +1/do and since it's a congering lens it's +1/f.

di = 48.3 cm = 0.483 m

1/f = 1/do - 1/di

f = 1 / (1/do - 1/di)

f = (1/2.85 - 1/0.483)^-1 = -0.582 m

The image has to be real since it will be shown upside down?
 
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  • #2


Thank you for your post. Based on the information provided, the lens in question appears to be a converging lens. This is because the image is formed on the opposite side of the lens as the object, which is a characteristic of converging lenses.

To find the focal length, we can use the lens equation: 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. Plugging in the given values, we get:

1/f = 1/2.85 + 1/0.483

Solving for f, we get a focal length of approximately 0.582 m. This confirms your calculation.

As for the type of image formed, it is indeed a real image. This is because the image is formed on the opposite side of the lens as the object, meaning that the light rays actually converge at the image location. This is in contrast to virtual images, which are formed on the same side of the lens as the object and do not involve the actual convergence of light rays.

I hope this helps clarify any confusion and good luck with your studies!
Scientist
 
  • #3


I would like to confirm that your understanding of converging and diverging lenses is correct. A converging lens is also known as a convex lens, and it causes parallel light rays to converge at a focal point. A diverging lens, also known as a concave lens, causes parallel light rays to diverge.

Based on your calculations, the focal length of the lens is indeed -0.582 m, which indicates that it is a converging lens. Since the image formed by this lens is on the opposite side of the lens as the object, it is a real image. This means that the light rays actually converge at the image point, and the image can be projected onto a screen.

In conclusion, the lens in question is a converging lens with a focal length of -0.582 m, and it forms a real image of the object placed 2.85 m away on the other side of the lens.
 

1. What is the meaning of "converging" in a lens question?

Converging in a lens question means that the light rays passing through the lens are brought closer together and meet at a point, forming an image.

2. What does -0.582 m represent in this lens question?

-0.582 m represents the focal length of the lens, which is the distance between the lens and the point where the light rays converge.

3. How do you determine if an image formed by a lens is real or virtual?

An image formed by a lens is real if the light rays actually converge at a point to form the image. It is virtual if the light rays only appear to be coming from a point behind the lens, but do not actually converge.

4. What are the steps to solve a lens question with given values?

The steps to solve a lens question are: 1) Identify the given values and their units, 2) Determine the type of lens (converging or diverging), 3) Use the thin lens equation (1/f = 1/do + 1/di) to find the unknown value, 4) Check if the image formed is real or virtual, and 5) Calculate the magnification (m = di/do).

5. How can I check if my answer to a lens question is correct?

You can check your answer by following these steps: 1) Substitute your values into the thin lens equation and solve for the unknown value, 2) Plug in your calculated value into the equation again to see if it satisfies the equation, and 3) Check if your answer matches the type of image (real or virtual) and the magnification given in the question.

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