Exercise-5a
Question 1. What is a lens?
Solution:
A lens is a transparent refracting medium that is surrounded by two usually spherical curved surfaces.
Question 2. Name the different kinds of lenses. Draw diagrams to illustrate them.
Solution:
There are two types of lens: –
(i) Convex or converging lens,
(ii) Concave or diverging lens.
Question 3. State two differences between a convex and a concave lens in their
(a) appearance, and (b) action on the incident light
Solution:
Question 4. Which lens is converging?
(i) An equiconcave lens or an equiconvex lens
(ii) A concavo-convex lens or a convexo-concave lens?
Solution:
(i) An equi-convex lens is convergent lens.
(ii) A converging lens is the concavo-convex lens. This is due to the fact that it is thicker in the center and thinner at the edges, allowing it to concentrate all incidents light.
Question 5. Out of the two lenses, one concave and the other convex, state which one will show the divergent action on a light beam.
Solution:
A light beam’s divergent movement will be visible through a concave lens.
Question 6. Show by a diagram the refraction of two light rays incident parallel to the principal axis on a convex lens by treating it as a combination of a glass block and two triangular glass prisms.
Solution:
The convex lens has one glass block and two glass prisms, as seen in the illustration. The glass block has one glass prism above it and another below it.
Question 7. Show by a diagram, the refraction of two light rays incident parallel to the principal axis on a concave lens by treating it as a combination of a glass block and two triangular glass prisms.
Solution:
The concave lens has one glass block and two glass prisms, as seen in the figure. The glass block has one glass prism above it and another below it.
Question 8. How does the action of a convex lens differ from that of a concave lens on a parallel beam of light incident on them? Draw diagram to illustrate your answer.
Solution:
The top portion of a convex lens bends the incident ray downward when a parallel beam of light strikes it. While the middle section of the light passes it without deviating, the bottom half bends the ray upward
However, in the case of a concave lens, the center component of the lens passes the ray undeviated while the top and lower portions of the lens bend the incident ray upwards and downwards, respectively.
Question 9. Define the term principal axis of a lens.
Solution:
It is the line that connects the two lens surfaces’ curvature centers.
Question 10. Explain optical center of a lens with the help of proper diagram(s).
Solution:
A beam of light travelling through this point on the lens’s main axis will emerge parallel to the direction in which it was incident. In the illustration, the letter O designates it. The optical center is thus the Centre of the lens.
Question 11. A ray of light incident at a point on the principal axis of a convex lens passes undeviated through the lens.
(a) What special name is given to this point on the principal axis?
(b) Draw a labeled diagram to support your answer in part (a)
Solution:
(a) This point is known as Optical Centre.
(b)
Question 12. State the condition when a lens is called an equi-convex or equi-concave
Solution:
When the radius of curvature of the two surfaces of a lens is identical, it is referred to as being equi-convex or equi-concave.
Question 13. Define the term principal foci of a convex lens and illustrate your answer with the aid of proper diagrams.
Solution:
Any direction of a light beam will pass through a lens. A lens therefore has two main focuses. The first focal point of a convex lens is a point F1 on its principal axis where, following refraction through the lens, light rays emanating from or passing through it become parallel to the lens’s principal axis. A position F2 on the primary axis that, after the lens has refracted light rays incident parallel to it, allows them to pass through is the second focal point for a convex lens.
Question 14. Define the term principal foci of a concave lens and show them with the help of proper diagrams.
Solution:
Any direction of a light beam will pass through a lens. A lens therefore has two main focuses.
The first focal point of a concave lens is a point F1 on its principal axis where incident light seems to meet after being refracted by the lens and being parallel to its principal axis. A concave lens’s second focal point is a position F2 on its primary axis where, upon refraction by the lens, light rays incident parallel to the principal axis seem to be diverging.
Question 15. Draw a diagram to represent the second focus of a concave lens.
Solution:
The above diagram shows the second focus represented by a concave lens.
Question 16. Draw a diagram to represent the second focus of a convex lens.
Solution:
The above diagram shows the second focus represented by a convex lens.
Question 17. A ray of light, after refraction through a concave lens emerges parallel to the principal axis. (a) Draw a ray diagram to show the incident ray and its corresponding emergent ray. (b) The incident ray when produces meets the principal axis at a point. Name the point.
Solution:
(a)
(b) The initial focus is the location where the incident ray meets the primary axis when it is created.
Question 18. A ray of light after refraction through a convex lens emerges parallel to principal axis.
(a) Draw a ray diagram to show it.
(b) The incident ray passes through a point on the principal axis. Name the point
Solution:
(a)
(b) The initial focus is the location where the incident ray passes through a point on the primary axis.
Question 19. A beam of light incident on a convex lens parallel to its principal axis converges at a point on the principal axis. Name the point. Draw a ray diagram to show it.
Solution:
A beam of light incident on a convex lens parallel to its principal axis converges at a point on the principal axis the point will be the secondary focus.
Question 20. A beam of light incident on a thin concave lens parallel to its principal axis diverges and appears to come from a point on the principal axis. Name the point. Draw a ray diagram to show it.
Solution:
A beam of light incident on a thin concave lens parallel to its principal axis diverges and appears to come from a point on the principal axis, which seems to originate from “Second Focus.”.
Question 21. Define the term focal length of a lens.
Solution:
The focal length of a lens is defined as the distance between its optical Centers, abbreviated O, and its secondary focal point.
Question 22. What do you mean by the focal plane of a lens?
Solution:
A first focal plane is a plane that is perpendicular to the lens’s main axis and passes through the lens’ focal point.
Question 23. State the condition for the following
(i) A lens has both its focal lengths equal
(ii) A ray passes undeviated through the lens
Solution:
(i) If a lens has both its focal length equal medium is the same on either side of the lens.
(ii) If a ray passes undeviated through the lens it is incident at the optical center of the lens.
Question 24. A parallel oblique beam of light falls on a
(i) Convex lens,
(ii) Concave lens. Draw a diagram in each case to show the refraction of light through the lens.
Solution:
Refraction of an oblique parallel beam by a convex lens.
Refraction of an oblique parallel beam by a concave lens
Question 25. (i) The diagram alongside shows a lens as a combination of a glass block and two prisms Name the lens formed by the combination.
(ii) The diagram alongside shows a lens as a combination of a glass block and two prisms. What is the rays the line XX’ called?
(iii) The diagram alongside shows a lens as a combination of a glass block and two prisms. Complete the ray diagram and show the path of the incident ray AB after passing through the lens
(iv) The diagram alongside shows a lens as a combination of a glass block and two prisms. The final emergent ray will either meet XX’ at a point or appear to come from a point on XX’. Label the point as F. What is this point called?
Solution:
(i) The lens form by the combination is a convex lens.
(ii) The primary axis is XX’.
(iii) The full illustration is
(iv) The focal point or focus is designated as point F.
Question 26. The diagram alongside shows a lens as a combination of a glass block and two prisms
(i) Name the lens formed by the combination.
(ii) What is the line XX’ called?
(iii) Complete the path of the incident ray AB after passing through the lens.
(iv) The final emergent ray either meets XX’ at a point or appears to come from a point on XX’ Label it as F. what is this point called?
Solution:
(i) The combination forms concave lens.
(ii) XX’ is known as principal axis.
(iii) Complete diagram is drawn as
(iv) The point F is called as Focal point or focus.
Question 27. In following figure, F1 and F2 are the positions of the two foci of the thin lenses shown in diagram (a) and (b) draw accurately the path taken by the light ray AB after it emerges from the lens in each diagram (a) and (b).
Solution:
Question 28. In following figure, F1 and F2 are the two foci of the thin lenses shown in diagram (a) and (b) and AB is the incident ray. Complete the diagram to show the path of the ray AB after refraction through the lens in each diagram (a) and (b).
Solution:
Question 29. Complete the following sentence:
(i) If half part of a convex lens is covered, the focal length ……………… change, but the intensity of image ……………
(ii) A convex lens is placed in water. Its focal length will ……………………
(iii) The focal length of a then convex lens is …………….. than that of a thick convex lens.
Solution:
(i) If half part of a convex lens is covered, the focal length does not change, but the intensity of image decreases .
(ii) A convex lens is placed in water. Its focal length will increase .
(iii) The focal length of a thin convex lens is more than that of a thick convex lens.
Exercise-5a
Multiple Choice Types
Question 1. A ray of light after refraction through a lens emerges parallel to the principal axis of the lens. The incident ray either passes through:
a) its optical centre
b) its first focus
c) its second focus
d) its center of curvature of the first surface
Solution: b) its first focus
Question 2. A ray of light incident on a lens parallel to its principal axis, after refraction passes through or appears to come from:
a) Its first focus
b) Its optical entre
c) Its second focus
d) The center of curvature of its second surface
Solution: Its second focus
Exercise-5b
Question 1. What are the three principal rays that are drawn to construct the ray diagram for the image formed by a lens? Draw diagrams to support your answer.
Solution:
(i) A light beam incident at the optical Centre O of the lens travels through it without deviating.
(ii) In a convex lens, a light beam incident parallel to the primary axis travels through the second focus F2 after refraction or seems to emanate from the second focus F2 (in a concave lens).
(iii) After refraction, a light beam emerges parallel to the major axis after passing through the first focus F1 (in a convex lens) or aimed at the first focus F1 (in a concave lens).
Question 2. In the diagram below, XX’ represents the principal axis, O the optical center and F the focus of the lens. Complete the path of rays A and B as they emerge out of the lens.
Solution:
Question 3. Where must a point source of light be placed in front of a convex lens so as to obtain a parallel beam of light?
Solution:
When a point source of light is positioned at the convex lens’s “first focal point,” or the focus point to the left of the optical Centre, light rays are refracted via length to form a parallel beam of light.
Question 4. Distinguish between a real and a virtual image.
Solution:
Question 5. Study the diagram given below.
(i) Name the lens LL’
(ii) What are the points O and O’ called?
(iii) Complete the diagram to form the image of the object AB.
(iv) State the three characteristics of the image.
(v) Name a device in which this action of lens is used.
Solution:
(i) In the above diagram LL’ lens is convex lens.
(ii) The first and second focus points are O and O’, respectively.
(iii)
(iv) The created image will be:-
a) Enlarged
b) Virtual
c) Upright and beyond C
(v) A magnifying glass uses a lens with this kind of movement.
Question 6. Study the following diagram:
(i) Name the lens LL’.
(ii) What are the points O, O’ called?
(iii) Complete the diagram to form the image of the object AB.
(iv) State the three characteristics of the image.
Solution:
(i) The LL’ lens is concave.
(ii) The first and second focus points, respectively, are designated as O and O
(iii)
(iv) The picture has the following three qualities:
a) Virtual
b) Erect
c) Diminished
Question 7. The following diagram shows an object AB and a converging lens L with foci F1 and F2.
(i) Draw two rays from the object and complete the diagram to locate the position of the image. Mark the image CD. Clearly mark on the diagram the position of the eye from where the image can be viewed.
(ii) State three characteristics of the image in relation to the object.
Solution:
(i) The complete diagram is
(ii) The resulting picture will be enlarged, virtual, and upright.
Question 8. The diagram given below shows the position of an object OA in relation to a converging lens whose foci are at F1 and F2.
(i) Draw two rays to locate the position of the image.
(ii) State the position of image with reference to the lens.
(iii) Describe three characteristic of the image.
(iv) Describe how the distance of the image from the lens and the size of the image change as the object is moved towards F1.
Solution:
(i)
(ii) The pictures will be located at a distance more than double the lens’s focal length.
(iii) The picture will be actual, enlarged, and reversed
(iv) The picture will shift away from F2 and get enlarged as the item moves near F1. At F1, an extremely enlarged picture will develop at infinity. The picture will form on the same side of the item and be amplified between F1 and the optical center.
Question 9. A converging lens forms the image of an object placed in front of it beyond 2F2 of the lens.
(i) Where is the object placed?
(ii) Draw a ray diagram to show the formation of image.
(iii) State three characteristics of the image.
Solution:
(i) The object is placed beyond 2F1.
(ii)
(iii) The image obtain diminished, real and inverted.
Question 10. A convex lens forms an image of an object equal to the size of the object. (i) Where is the object placed in front of the lens?
(ii) Draw a diagram to illustrate it.
(iii) State two more characteristics of the image.
Solution:
(i) The item is positioned at the center of the curve.
(ii)
(iii) The image will formed is real and inverted.
Question 11. A lens forms an erect, magnified and virtual image of an object.
(i) Name the type of lens.
(ii) Where is the object placed in relation to the lens?
(iii) Draw a ray diagram to show the formation of image.
(iv) Name the device which uses this principle.
Solution:
(i) The given Lens is convex lens
(ii) It is positioned between the lens and the focus (F1).
(iii)
(iv) This idea is applied in “Magnifying glass.”
Question 12. A lens always forms an image between the object and the lens.
(i) Name the lens.
(ii) A lens always forms an image between the object and the lens.
(iii) State three characteristics of the image.
Solution:
(i) A concave lens creates the image between the object and itself.
(ii) Ray diagram
(iii) The image obtained is virtual, erect and diminished.
Question 13. Classify as real or virtual, the image of a candle flame formed on a screen by a convex lens. Draw a ray diagram to illustrate how the image is formed.
Solution:
Place the candle beyond 2F1 so that its actual and inverted, decreased image forms between F2 and 2F2. The candle in this instance is AB, and between F2 and 2F2, its real and inverted image is created.
Question 14. Show by a ray diagram that a diverging lens cannot form a real image of an object placed anywhere on its principal axis.
Solution:
Question 15. Draw a ray diagram to show how a converging lens can form a real and enlarged image of an object.
Solution:
The image obtained in above diagram is
a) Real
b) Enlarged
c) Inverted.
Question 16. A lens forms an upright and diminished image of an object irrespective of its position. What kind of lens is this?
Solution:
A Concave lens forms an upright and diminished image of an object irrespective of its position.
Question 17. Draw a ray diagram to show how a converging lens is used as a magnifying glass to observe a small object. Mark on your diagram the foci of the lens and the position of the eye.
Solution:
The item is positioned between the convex lens’s expanded side and focal point F1, where its image is created. Consequently, this lens has a magnifying function.
Question 18. Draw a ray diagram to show how a converging lens can form an image of the sun. Hence give a reason for the term ‘burning glass’ for a converging lens used in this manner.
Solution:
Since the sun is at infinity, a convex lens creates a true but greatly scaled-down image of it at the second focus point. The sun’s light, which is at infinity, is brought into focus on a sheet of paper held at the second focal plane of the lens by employing the convex lens as burning glass. The paper burns because the sun’s rays are hot enough. As a result, this lens is known as “burning glass.”
Question 19. A lens forms an inverted image of an object.
(i) What kind of lens is this?
(ii) What is the nature of the image real or virtual?
Solution:
(i) Convex lens forms an inverted image of an object.
(ii) The image is real in its nature.
Question 20. A lens forms an upright and magnified image of an object
(i) Name the lens.
(ii) State whether the image is real or virtual
Solution:
(i) Convex lens forms an upright and magnified image of an object.
(ii) Image formed-Virtual.
Question 21. Name the lens which always forms an erect and virtual image.
(a) State whether the image in part
(b) is magnified or diminished.
Solution:
(a) Concave lens always forms an erect and virtual image.
(b) Here the Image size of mage diminished.
Question 22. Can a concave lens form an image of size two times that of the object? Give reason?
Solution:
Because it constantly creates a smaller picture, a concave lens is unable to create an image that is twice as large as the item.
Question 23. Give two characteristic of the image formed by a concave lens.
Solution:
A concave lens creates a virtual and weaker image.
Question 24. Give two characteristic of the virtual image formed by a convex lens.
Solution:
A convex lens will provide an upright, enlarged virtual picture.
Question 25. In the following cases, where must an object be placed in front of a convex lens so that
(i) The image formed is at infinity.
(ii) The image formed is of same size as the object
(iii) The image formed is inverted and enlarged
(iv) The image formed is upright and enlarged image?
Solution:
(i) at focus.
(ii) At 2F.
(iii) Between F and 2F.
(iv) Between optical center and focus.
Question 26. Complete the following table:
Solution:
Question 27. State the changes in the position, size and nature of the image of an object when brought from infinity up to a convex lens. Illustrate your answer by drawing ray diagrams.
Solution:
(i) When the item is at infinity, the picture is at F2, it is actual and inverted, and it is much shrunk in size.
(ii) When the object (AB) is located beyond 2F1, the picture (A’B’) is smaller, less real, and inverted when it is between F2 and 2F2.
(iii) When the item (AB) is located at 2F1, the picture (A’B’) is located at 2F2, and it is both actual and inverted, as well as the same size as the thing.
(iv) When the item (AB) is positioned between 2F1 and F1, the picture (A’B’) is located beyond 2F2 and is actual, enlarged, and inverted.
(v) When the item (AB) is positioned at F1, the image’s location is at infinity; it is much enlarged, real, and upside-down.
(vi) When the object (AB) is between the lens and F1, the image (CD) is placed behind the object on the same side; it is enlarged, virtual, and upright.
Question 28. State the changes in the position, size and nature of the image of an object when brought from infinity up to a concave lens. Illustrate your answer by drawing ray diagrams.
Solution:
(i) Parallel rays from object (AB) appear to fall on concave lens when it is at infinity. Picture is formed at focus as a result. This image is virtually shown, greatly shrunk, and upright.
(ii) An image forms between the focus and the optical center of the lens when object (AB) is positioned at any location between infinity and the optical center of the lens. This picture is upright, virtual, and scaled down.
Question 29. Complete the following sentence:
(i) An object is placed at a distance of more than 40 cm from a convex lens of focal length 20 cm. The image formed is real, inverted and ……….
(ii) An object is placed at a distance 2f from a convex lens of focal length f. The size of image formed is ………….. That of the object
(iii) An object is placed at a distance 5 cm from a convex lens of focal length 10 cm. The image formed is virtual, upright and …………….
Solution:
(i) An object is placed at a distance of more than 40 cm from a convex lens of focal length 20 cm. The image formed is real, inverted and Diminished.
(ii) An object is placed at a distance 2f from a convex lens of focal length f. The size of image formed is Equal That of the object
(iii) An object is placed at a distance 5 cm from a convex lens of focal length 10 cm. The image formed is virtual, upright and Magnified .
Question 30. State whether the following statement are ‘true’ or ‘false’ by writing T/F against them.
(i) A convex lens has a divergent action and a concave lens has a convergent action.
(ii) A concave lens, if kept at a proper distance from an object, can form its real image.
(iii) A ray of light incident parallel to the principal axis of a lens passes undeviated after refraction.
(iv) A ray of light incident at the optical center of lens, passes undeviated after refraction.
(v) A concave lens forms a magnified or diminished image depending on the distance of object from it.
Solution:
Question 31. Multiple Choice Types:
Question 1. For an object placed at distance 20 cm in front of a convex lens, the image is at distance 20 cm behind the lens. The focal length of convex lens is:
a) 20 cm
b) 10 cm
c) 15 cm
d) 40 cm
Solution: 10 cm
Question 2. For the object placed between optical centre and focus of a convex lens, the image is:
a) Real and enlarged
b) Real and diminished
c) Virtual and enlarged
d) Virtual and diminished
Solution: Virtual and enlarged
Question 3. A concave lens forms the image of an object which is:
a) Virtual, inverted and diminished
b) Virtual, upright and diminished
c) Virtual, inverted and enlarged
d) Virtual, upright and enlarged
Solution: Virtual, upright and diminished
Exercise-5c
Question 1. State the sign convention to measure the distances for a lens.
Solution:
The primary axis is the axis along which distances are measured. These distances are gauged from the lens’ optical center. All distances that are measured in the same direction as the light’s incoming beam are considered positive, whilst distances measured in the opposite direction are considered negative. The lengths above the major axis are all regarded as positive, whilst the lengths below the principal axis are regarded as negative. Convex lenses are given positive focus lengths, whereas concave lenses are given negative focal lengths.
Question 2. The focal length of a lens is (i) positive, (ii) negative.
In each case, state the kind of lens.
Solution:
(i) A lens is said to be convex if its focal length is positive.
(ii) A lens is considered concave if its focal length is negative.
Question 3. Write the lens formula explaining the meaning of the symbols used.
Solution:
The Lens formula is:-
u = The object distance is the separation of the object from the optical center.
v = The image distance is the separation of the picture from the optical center.
f = The distance of the principal focus from the optical center is called the focal length.
Question 4. What do you understand by the term magnification? Write expression for it for a lens, explaining the meaning of the symbols used.
Solution:
The term magnification means a comparison between the sizes of the image formed by a lens with respect to the size of the object. For a lens: Magnification ‘m’ is the ratio of the height of the image to the height of the object.
Question 5. What information about the nature of image (i) real or virtual, (ii) erect or inverted, do you get from the sign of magnification + or – ?
Solution:
(i) A positive magnification sign denotes a virtual picture, whereas a negative sign denotes a real image.
(ii) A positive magnification sign denotes an upright picture, whereas a negative sign denotes an inverted image.
Question 6. Define the term power of a lens. In what unit is it expressed?
Solution:
The power of a lens is a measurement of the deviation in the rays’ paths that it causes when they are refracted through it. Diopter is its measure (D).
Question 7. How is the power of a lens related to its focal length?
Solution:
Question 8. How does the power of a lens change is its focal length is doubled?
Solution:
A lens’s power is half as the focal length doubles.
Question 9. How is the sign (+ or -) of power of a lens related to its divergent or convergent action?
Solution:
The direction in which a light ray is bent by the lens determines the power sign. Both positive and negative powers are possible. The power of a lens is positive when a ray diverges away from its center, whereas the power is negative when the beam diverges towards it.
Question 10. The power of a lens is negative. State whether it is convex or concave?
Solution:
Concave lens has negative power.
Question 11. Which lens has more power: a thick lens or a thin lens?
Solution:
A thick lens has more power than a thin lens. This is because a thick lens has a larger surface curvature. It has a short focal length and that is why it deviates the rays of light at a greater extent.
Question 12. Describe how you would determine the focal length of a converging lens, using a plane mirror and one pin. Draw a ray diagram to illustrate your answer.
Solution:
We require a vertical stand, a plane mirror, a lens, and a pin to calculate focal length using a plane mirror. Placing the lens L on a horizontal plane mirror MM’. Place a pin P such that its point is vertically above the lens’s center O on the clamp of a vertical stand.
As seen from vertically above the pin, adjust the height of the pin until there is no parallax (i.e., when the pin and its image shift together). Now, using a metre scale and a plumb line, measure the distances of the pin from the lens and the mirror, respectively Calculate the average distance between the two. This displays the lens’s focal length, i.e.,
Question 13. State two applications of a
(i) Convex lens.
(ii) Concave lens
Solution:
(i) The two applications of a convex lens are:-
a) It is used as an objective lens in a telescope, camera, slide projector, etc.
b) With its short focal length, it is also used as a magnifying glass.
(ii) The two applications of a concave lens are:-
a) A person suffering from short-sightedness or myopia wears spectacles having a concave lens.
b) A concave lens is used as an eye lens in a Galilean telescope to obtain an erect final image of the object.
Question 14. How will you differentiate between a convex and a concave lens by looking at?
(i) A distant object,
(ii) A printed page
Solution:
(i) If an object’s inverted image is visible when viewed through a lens, the lens is convex; if its upright image is visible, the lens is concave.
(ii) If the letters on a printed page look enlarged when the lens is kept close to it, the lens is convex; if the letters appear decreased, the lens is concave.
Question 15. Multiple choice types:-
1. If the magnification produced by a lens is – 0.5, the correct statement is:
a) The lens is concave
b) The image is virtual
c) The image is magnified
d) The images is real and diminished formed by a convex
Solution: d) The images is real and diminished formed by a convex
2. The correct lens formula is
Solution:
3. On reducing the focal length of a lens, its power:
a) Decreases
b) Increases
c) Does not change
d) First increases then decreases
Solution: b) Increases
4. The power of a lens is + 1.0 D is :
a) convex of focal length 1.0 cm
b) convex of focal length 1.0 m
c) concave of focal length 1.0 cm
d) concave of focal length 1.0 m
Solution: b) Convex of focal length 1.0 m.
Numerical
Question 1. At what position a candle of length 3 cm be placed in front of a convex lens so that its image of length 6 cm be obtained on a screen placed at distance 30 cm behind the lens? What is the focal length of lens in part (a)?
Solution:
Height of the candle (object) = 3 cm
Height of the image of the candle = 6 cm
Image distance = 30 cm
(a) The formula for magnification of a lens is
Question 2. A concave lens forms the image of an object kept at a distance 20 cm in front of it, at a distance 10 cm on the side of the object.
What is the nature of the image?
Find the focal length of the lens.
Solution:
Object distance, u = -20 cm
Image distance, v = -10 cm
(a) The image is formed on the same side as the object. Hence, it is a virtual image. Also, since the lens is a concave lens the image will be erect and diminished.
(b) Lens formula is
Question 3. The focal length of a convex lens is 25 cm. At what distance from the optical center of the lens an object be placed to obtain a virtual image of twice the size?
Solution:
Focal length, f = + 25cm
Image is virtual and magnified, m = + 2
For a lens, magnification is
Question 4. Where an object should be placed in front of a convex lens of focal length 0.12 m to obtain a real image of size three times the size of the object, on the screen?
Solution:
Focal length of a convex lens, f = + 0.12m
m=-3 (real image)
For a lens, magnification is
Question 5. An illuminated object lies at a distance 1.0 m from a screen. A convex lens is used to form the image of object on a screen placed at distance 75 cm from the lens . Find: (i) the focal length of lens, and (ii) the magnification.
Solution:
Image distance, v = 75 cm
Object distance, u = -25 cm
Lens formula is,
Question 6. A lens forms the image of an object placed at a distance 15 cm from it, at a distance 60 cm in front of it. Find:
(i) The focal length
(ii) The magnification
(iii) The nature of image
Solution:
(i) Object distance, u = -15 cm
Image distance, v = -60 cm
Lens formula is,
Question 7. A lens forms the image of an object placed at a distance of 45 cm from it on a screen placed at a distance 90 cm on other side of it. Find the
(a) Name the kind of lens.
(b) (i) the focal length of lens
(ii) the magnification of image.
Solution:
(a) Object distance, u = -45 cm
Image distance, v = + 90 cm
As the image is formed on the other side of the lens, the image is real. Hence, the lens is a convex lens.
(b) Object distance, u = -45 cm
Image distance, v = + 90 cm
(i) Lens formula is,
Question 8. An object is placed at a distance of 20 cm in front of a concave lens of focal length 20 cm. find:
(i) The position of image.
(ii) The magnification of image.
Solution:
(i) Object distance, u = -20 cm
Focal length, f = -20 cm (concave lens)
Lens formula is,
v = -10 cm
Hence, the image is 10 cm in front of the lens on the same side as the object.
(ii) Object distance, u = -20 cm
Focal length, f = -20 cm (concave lens)
For a lens, magnification is
Question 9. A convex lens forms an inverted image of size same as that of the object which is placed at a distance 60 cm in front of the lens. Find:
(i) The position of image
(ii) The focal length of the lens
Solution:
(i) When an item is positioned at 2f, i.e. (u = 2f), a convex lens creates an inverted, actual, and identically sized picture of the object.
In these circumstances, the focal length of the picture is doubled at the lens’s opposite side, where it forms (2f2).
(ii) To find the focal length of this lens, we use the relationship:
Object distance (u) = 2f
It is given that,
Object distance = 60 cm
60 = 2f
f = 30 cm
So, the focal length of this lens is 30 cm.
Question 10. A concave lens forms an erect image of 1/3rd size of the object which is placed at a distance 30 cm in front of the lens. Find:
(i) The position of image
(ii) The focal length of the lens.
Solution:
(i) Given that the concave lens forms an erect image of 1/3rd size of the object.
That implies that the magnification provided by the lens is 1/3.
Magnification is given by
The position of the image formed is 10 cm on the same side of the lens where the object is placed.
(ii) Focal length of a lens is given by
The focal length of the given lens is 15 cm.
Question 11. The power of a lens is +2.0 D. Find its focal length and state what kind of lens it is?
Solution:
P = + 2.0 D
Question 12. Express the power (with sign) of a concave lens of focal length 20 cm.
Solution:
Question 13. The focal length of a convex lens is 25 cm. Express its power with sign.
Solution:
Focal length, f = +25 cm = +0.25 m
Power of a lens is
Question 14. The power of a lens is -2.0 D. Find its focal length and its kind.
Solution:
Question 15. The magnification by a lens is -3. Name the lens and state how are u and v related?
Solution:
The fact that the magnification value is negative indicates that the image is real and inverted. When magnification is more than 1, the size of the picture is increased. Consequently, a convex lens should be used.
Relation between u and v is given by
Question 16. The magnification by a lens is +0.5. Name the lens and state how are u and v related?
Solution:
The image formed by the concave lens is always virtual, erect and smaller than the object. Therefore the magnification is always positive and less than 1.
Question 17. A concave lens is a focal length 30 cm. Find the position and magnification (m) of image for an object placed in front of it at distance 30 cm. State whether the image is real on virtual?
Solution:
Object distance = -30cm
Focal length = f = -3-cm
Image distance = v = ?
Question 18. Find the position and magnification of the image of an object placed at distance of 8.0 cm in front of a convex lens of focal length 10.0 cm. Is the image erect or inverted?
Solution:
Object distance = u = -8cm
Focal length f = 10cm
Image distance v = ?
Exercise-5d
Question 1. What is magnifying glass? State its two uses.
Solution:
A convex lens with a restricted field of view is a magnifying glass. For ease of usage, it is fitted in a lens holder. It is employed to read and see the tiny numbers and letters. Watchmakers use it to observe the watch’s tiny components and screws.
Question 2. Draw a neat labeled ray diagram to show the formation of image by a magnifying glass. State three characteristics of the image
Solution:
If an item (AB) is placed between a convex lens’ focal length and optical center, the image of that object (A’B’) will form on the opposite side of the Lens.
The image obtained will be virtual, magnified and erect.
Question 3. Where is the object placed in reference to the principal focus of a magnifying glass, so as to see its enlarged image? Where is the image obtained?
Solution:
Between the lens and the principal focus is the object. Between the lens and the principal focus is where the picture is obtained.
Question 4. Define magnifying power of a simple microscope. How can it be increased?
Solution:
The ratio of the angle subtended by the image at the eye to the angle subtended by the object (assumed to be positioned at the least distance of distinct vision D = 25 cm) at the eye is the definition of the magnifying power of the microscope, i.e.
Magnifying power = 1 + D/F
Where F is the focal length of the lens. The magnifying power of a microscope can be increased by using the lens of short focal length. But it cannot be increased indefinitely.
Question 5. State two applications each of a convex lens and concave lens
Solution:
The two applications of a convex lens are:-
(i) In a telescope, camera, slide projector, etc., it serves as the objective lens.
(ii) It may also be used as a magnifying glass due to its small focus length.
The two applications of a concave lens are:-
(i) Someone with myopia or short-sightedness uses glasses with concave lenses.
(ii)To create an upright final picture of the object, a concave lens is utilized as the eye lens in a Galilean telescope.
Question 6. Describe in brief how you would determine the approximate focal length of a convex lens.
Solution:
The idea that a beam of parallel rays incident from a distant object is converging on the focal plane of the lens may be used to estimate the approximate focal length of a convex lens.
A metre scale is set horizontally with its 0 cm end contacting the wall in an open area and next to a white wall.
Focus a far-off item on a wall by adjusting the convex lens up and down along the scale. The picture that develops on the wall is extremely close to the lens’ focus, and the metre scale may be used to read how far away the lens is from the image. This approximates the lens’s focal length.
Question 7. The following diagram shows the experimental set up for the determination of focal length of a lens using a plane mirror.
(i) Draw two rays from the point O of the object to show the formation of image I at O itself.
(ii) What is the size of the image I?
(iii) State two more characteristics of the image I.
(iv) Name the distance of the objects O from the optical center of the lens.
(v) To what point will the rays return if the mirror is moved away from the lens by a distance equal to the focal length of the lens?
Solution:
(i)
(ii) The image’s size will match that of the object.
(iii) The resulting image will be actual and upside-down.
(iv) The focal length of the lens will be equal to the distance between object O and the optical lens.
(v) As long as the lens’ rays fall normally on the plane mirror M, the position of the mirror from the lens has no impact on how the picture is formed.
Question 8. Describe how you would determine the focal length of a converging lens, using plane mirror and one pin. Draw a ray diagram to illustrate your answer.
Solution:
We require a vertical stand, a plane mirror, a lens, and a pin to calculate focal length using a plane mirror. Placing the lens L on a horizontal plane mirror MM’. Place a pin P such that its point is vertically above the lens’s center O on the clamp of a vertical stand.
As seen from vertically above the pin, adjust the height of the pin until there is no parallax (i.e., when the pin and its image shift together).
Now, using a metre scale and a plumb line, measure the distances of the pin from the lens and the mirror, respectively Calculate the average distance between the two This provides the lens’s focal length i.e., F = x+y/2
Question 9. How will you differentiate between a convex and a concave lens by looking at a
(i) Distant object
(ii) Printed page?
Solution:
(i) If an object’s inverted image is visible when viewed through a lens, the lens is convex; if its upright image is visible, the lens is concave.
(ii) If the letters on a printed page look enlarged when the lens is kept close to it, the lens is convex; if the letters appear decreased, the lens is concave.
Question 10. Multiple Choice Types
Q1. A magnifying glass forms:
a) A real and diminished image
b) A real and magnified image
c) A virtual and magnified image
d) A virtual and diminished image
Solution: c) A virtual and magnified image
Question 2. The maximum magnifying power of a convex lens of focal length 5 cm can be:
a) 25
b) 10
c) 1
d) 6
Solution: d) 6
Focal length of convex lens, f=+5 cm
Magnifying power of convex lens (simple microscope) is
