# Bài tập Ảnh của một vật tạo bởi thấu kính hội tụ

Exercise Image of an object formed by a converging lens is one of the important knowledge that helps 8th graders to solve various types of Physics exercises. So how to construct an image of an object formed by a converging lens? Invite 9th grade students to follow Download.vn to follow the article below.

The image of an object created by a converging lens summarizes the theory with several types of multiple-choice exercises and essay questions with answers. Thereby helping students consolidate and master the foundational knowledge, apply them with basic exercises to achieve high results in the upcoming exam. So here is the detailed content of the document, please follow along here.

## Exercise Image of an object formed by a converging lens

### I. Image characteristics of an object formed by a converging lens

– Objects placed out of focus range for real images opposite to the object. When the object is placed very far from the lens, the real image is located at a distance equal to the focal length of the lens.

– Objects placed in the focal range give a virtual image larger than the object and in the same direction as the object.

Attention:

+ Virtual images are not displayed on the screen but can be seen by the eye when the eye is placed behind the lens to receive the peeking beam. + The real image can be clearly displayed on the screen or seen by the eye when the eye is placed behind the focus point of the emerging beam. ### II. How to construct an image of an object formed by a converging lens

a) How to construct an image of bright spot S produced by a converging lens

– From S we construct two rays (of three special rays) to the lens, then draw two rays emerging from the lens.

– If the two emergent rays actually intersect, then the intersection point is the real image S’ of S, if the two emergent rays do not actually intersect but have their extension lines intersect, then that intersection point is exactly. is the virtual image S’ of S through the lens. b) Construct an image of bright object AB formed by a converging lens

To construct the image A’B’ of AB through the lens (AB is perpendicular to the lens, A lies on the principal axis), simply construct the image B’ of B with two of the three special rays, then from B’ lower perpendicular to the main axis, we have the image A’ of A. Attention: When rendering, the virtual image and the extension of the light ray are drawn with dashed lines

### III. Method of solving the image of an object formed by a converging lens

1. How to determine the position of an image when knowing the position of the object and the focal length, or determine the position of the object when the position of the image and the focal length is known, or determine the focal length when the position of the image and the position of the object is known .

Method 1: Draw an image of an object using the above method. Use properties of similar triangles to deduce the quantity to be determined.

Method 2: Applying formula to determine.

In which: the object is the real thing.

• f is the focal length of the lens (which is the distance from the focal point to the optical center).
• d is the distance from the position of the object to the lens.
• d’ is the distance from the position of the image to the lens (when real image d’ > 0, when virtual image d’ < 0).
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### IV. Determine the height of the object or the image

Method 1: Apply properties of similar triangles.

Method 2: Applying formula Where: h and h’ are the heights of the object and the image (when the image is real, h’ > 0, when it’s virtual, h’ < 0).

### V. Exercise Image of an object formed by a converging lens

Question 1: Object AB placed in front of the converging lens gives image A’B’, image and object lie on the same side with respect to the lens. Photo A’B’

A. is a real image, larger than the object.

B. is a virtual image, smaller than the object.

C. opposite to the object.

D. is a virtual image, in the same direction as the object. Image and object are on the same side with respect to the lens image A’B’ is virtual, in the same direction as the object

Verse 2: Image A’B’ of a bright object AB placed perpendicular to the principal axis at A and within the focal length of a converging lens is:

 A. virtual image inverted object. B. virtual image in the same dimension. C. real image in the same direction as the object. D. the real image is inverted.

When the object is placed in the focal range ⇒ the image is virtual, in the same direction as the object, larger than the object

Question 3: Object AB placed in front of the converging lens gives image A’B’, the image and the object lie on both sides of the lens, the image is:

 A. real, opposite to the object. B. real, always larger than the object. C. virtual, in the same direction as the object. D. real, always as tall as the object.

The image and the object are on both sides for the lens ⇒ the image is a real image, opposite to the object, can be smaller or equal to or larger than the object depending on the position of the object

Verse 4: C shows the wrong answer. Place a candle in front of a converging lens.

A. We can get the image of the candle on the screen.

B. The image of the candle on the screen may be larger or smaller than the candle.

C. The image of the candle on the screen can be a real image or a virtual image.

D. The virtual image of the candle is always larger than the candle.

The virtual image cannot be captured on the screen

Question 5: Placing an arrow-shaped object AB perpendicular to the principal axis of a converging lens of focal length f and a distance d = 2f from the lens, the image A’B’ of AB through the lens has the property:

 A. The image is real, in the same dimension and smaller than the object. B. the image is real, inverted and larger than the object. C. the image is real, inverted and smaller than the object. D. the image is real, inverted and as large as the object.

The image A’B’ of AB through the lens is real, inverted, and as large as the object.

Question 6: An object AB 3 cm high is placed in front of a converging lens. We get an image 4.5cm high. That photo is:

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 A. Real photo B. Virtual image C. Can be real or virtual D. Same dimension

In a converging lens, if the image size is larger than the object size, the image can be real or virtual

Verse 7: An object AB 2 cm high is placed in front of a converging lens and 10 cm from the lens. Using a screen M, we obtain an image A’B’ 4cm high as shown in the figure. The screen is a distance from the lens:

 A. 20cm B. 10cm C. 5cm D. 15 cm

A ray incident through optical center O gives a straight-line burst ray A and A’ lying on the same line through O ⇒

ΔABO ∼ ΔA’B’O OB/OB’ = AB/A’B’ = 2/4 = 1/2

⇒ OB’ = 2BO = 2.10 = 20 cm

So the screen is a distance OB’ = 20 cm . from the lens

Verse 8: An object AB is placed in front of a converging lens. Using a screen M, we obtain an image 5cm high and symmetric about the object through the optical center O. The dimensions of object AB are:

 A. 10cm B. 15cm C. 5 cm D. 20 cm

If the image is symmetric about the object through the optical center O, then the size of the object is equal to the size of the image:

⇒ AB = A’B’ = 5 cm

Verse 9: Given a lens of focal length 20 cm, object AB is placed 60 cm from the lens and has a height h = 2 cm.

a) Draw an image of the object through the lens.

b) Using geometric knowledge, calculate the distance from the image to the lens and the height of the image.

a) Image A’B’ is represented as shown in the figure: b. Let OA = d, OA’ = d’, OF = OF’ = f

We have: ΔABO ∼ A’OB so (first)
We have: ΔIOF’ B’OB’ so (2)
From (1) and (2) nice ⇒ fd’ = dd’ – fd

Divide both sides by d.d’.f, we get: From (1) we have: With f = 20cm, d = 60cm, then (cm)
So Question 10: A bright object AB has the form of a line segment perpendicular to the principal axis of a converging lens at A and 20 cm from the lens. The focal length of the lens is 15 cm.

a) Using special rays of light through the lens draw the image A’B’ of AB in the correct proportions.

b) Based on measurements and geometric knowledge, calculate how many times the image is higher than the object.

a. Picture: b. We have: ΔA’B’O ∼ ΔAOB, so we can infer the ratio: (first)
We have: ΔA’B’F’ ∼ ΔOIF’ so infer the ratio: (2)
Otherwise AB = OI (3)
From (1), (2), (3) deduce Substituting A’O = 60cm into expression (1) we get: So image A’B’ is 3 times taller than object AB.

Question 11: What does eye regulation do?

– Accommodation is the action of the eye that changes the focal length of the eye so that images of objects at different distances from the eye are still produced in the retina. This is done by the sphincter muscles of the eye.

When squeezed, these muscles cause the lens to swell, reducing the radius of curvature, so the focal length of the eye is reduced.

+ When the eye is in the unaccommodating state, the focal length of the eye is largest.

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When the eye muscles are maximally squeezed, the eye is in maximum accommodation and the focal length of the eye is minimal.

Verse 12

A bright object AB is placed perpendicular to the principal axis of a converging lens of focal length f = 12cm. Point A lies on the main axis. Construct image A’B’ of AB and comment on the characteristics of image A’B’ in two cases:

– Object AB is a distance d = 36cm from the lens (Figure 43.4a textbook).

– Object AB is at a distance d = 8cm from the lens (Figure 43.4b in textbook).  Construct an image of a bright object AB through a converging lens. Use two of the three learned light rays to construct the image B’ of point B.

+ Object AB is d = 36 cm from the lens, the object is outside the OF range.

The ray BI goes parallel to the principal axis, so the visible ray passes through F’.

The ray incident on BO is the ray passing through the center of O, so it lets the ray go straight

The two rays above intersect at B’, we get the real image B’ of B through the lens.

From B’ lower perpendicular to the axis of the lens, intersecting the principal axis at point A’. A’ is the image of point A. A’B’ is the image of AB formed by the converging lens. Comment: Image A’B” is a real image that is inverted to the object when the object is placed out of focus (Figure 43.4a).

+ Object AB is d = 8 cm from the lens, the object is in the OF range.

The ray BI goes parallel to the principal axis, so the visible ray passes through F’.

The ray incident on BO is the ray passing through the center of O, so it lets the ray go straight

The two rays above have the extension line intersecting at B’, we get the virtual image B’ of B through the lens.

From B’ lower perpendicular to the axis of the lens, intersecting the principal axis at point A’. A’ is the image of point A. A’B’ is the image of AB formed by the converging lens. Comment: The virtual image A’B’ is in the same direction as the object and larger than the object when the object is placed in the focal range (Figure 43.4b).

Verse 13:

Using knowledge of geometry, calculate the distance from the image to the lens and the height of the image in the two cases. Assume that object AB has height h = lcm.

We set: OA = d; OA’ = d’; OF = OF’ = f

+) Object AB is d = 36 cm from the lens, the object is outside the OF range. AB = h = 1cm

OA = d = 36cm

OF = OF’ = f = 12cm

A’O = ? A’B’ = ?

We have: We have: Which: OI = AB (3)

From (1), (2) and (3)  Substituting A’O = 18cm in (1) we have: So the height of the image is 0.5cm, the distance from the image to the lens is 18cm.

– TH2: Object AB is at a distance d = 8cm from the lens.

AB = h = 1cm

OA = d = 8cm

OF = OF’ = f = 12cm

A’O = ? A’B’ = ?

We have: We have: Which: OI = AB (3)

From (1), (2) and (3)  Substituting A’O = 24cm in (1) we have: So the height of the image is 3cm, the distance from the image to the lens is 24cm.

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