NCERT Exemplar Solution Class 10 Science Chapter 10
NCERT Exemplar Solution Class 10 Science Chapter 10 Light- Reflection and refraction all Questions Answer Solution. NCERT Exemplar Class 10 Science Chapter 10 Solution PDF.
NCERT Exemplar Solution Class 10 Science Chapter 10: Overview
NCERT Exemplar Solution Class 10 Science Chapter 10 |
|
Board |
NCERT |
Topic |
Exemplar Problem Solution |
Class |
10 |
Subject |
Science |
Chapter |
10 |
Chapter Name |
Light- Reflection and refraction |
NCERT Exemplar Solution Class 10 Science Chapter 10 Light- Reflection and refraction
Multiple Choice Questions:
1) Which of the following can make a parallel beam of light when light from a point source is incident on it?
(a) Concave mirror as well as convex lens
(b) Convex mirror as well as concave lens
(c) Two plane mirrors placed at 90° to each other
(d) Concave mirror as well as concave lens
Ans.(a) Concave mirror as well as convex lens.
Since, when a ray of light from point source passing through focal point of a concave mirror will be reflected back in such way that it must be parallel to principal axis of a concave mirror as shown in fig.10.1 a)
Also, when a ray of light from a point source is passed through the focal point of a convex lens it will emerge out as a ray of light which is parallel to principal axis of convex lens as shown in fig.10.1 b)
Fig.10.1 a) Fig.10.1 b)
2) A 10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is
(a) – 30 cm
(b) – 20 cm
(c) – 40 cm
(d) – 60 cm
Ans. (b) – 20 cm
Object height= 10mm=1cm
Image height= 5mm= 0.5cm
Image distance= -30cm (Since, if the image is real then image distance is taken as negative.)
Then, magnification (m) = image height/ object height = -v/u
0.5/1=-30/u
u = -30/0.5
u = – 60 cm
By Mirror formula,
1/f = 1/v + 1/u
= (1/-30) + (1/-60)
= (-2-1)/60
1/f = -1/20
Hence, f = -20 cm
Thus, the focal length of concave mirror is -20 cm.
Since, focal length of concave mirror is always negative.
3) Under which of the following conditions a concave mirror can form an image larger than the actual object?
(a) When the object is kept at a distance equal to its radius of curvature
(b) When object is kept at a distance less than its focal length
(c) When object is placed between the focus and center of curvature
(d) When object is kept at a distance greater than its radius of curvature
Ans. (c) When object is placed between the focus and center of curvature
In case of concave mirror, when the object MN is kept in between focus F and center of curvature C, then the image is formed beyond C which is enlarged, real and inverted as shown in fig, 10.3
Fig.10.3
4) Figure 10.1 shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is
(a) √3/ √2
(b) √2/ √3
(c) 1/ √2
(d) √2
Ans. (a) √3/ √2
Angle of incidence= 600
Angle of refraction= 450
Then, refractive index of medium B with respect to medium A is given by,
μBA = sini/sinr
= sin60/sin45
= (√3/2)/ (1/√2)
μBA = √3/√2
5) A light ray enters from medium A to medium B as shown in Figure 10.2. The refractive index of medium B relative to A will be
(a) Greater than unity
(b) less than unity
(c) Equal to unity
(d) Zero
Ans.(a) Greater than unity
Since, refracted ray of light in medium B bends towards normal, so medium B is more denser than medium A. hence speed of light VB in medium B is less than speed of light VA in medium A i.e. VB< VA.
Thus, refractive index of medium B with respect to medium A will be μBA= speed of light in medium A / speed of light in medium B
= VA/VBFig.10.5
Since, VB< VA
μBA> 1
6) Beams of light are incident through the holes A and B and emerge out of box through the holes C and D respectively as shown in the Figure10.3. Which of the following could be inside the box?
(a) A rectangular glass slab
(b) A convex lens
(c) A concave lens
(d) A prism
Ans. (a) A rectangular glass slab Because, we know that, in case of refraction through rectangular glass slab, the extent of bending of ray at a point of incidence is equal to the extent of bending of ray at a point of emergence. Due to which the incident ray and emergent ray will be parallel to each other as shown in fig.10.6 In other options this will not be happen, so the correct answer is rectangular glass slab only.
Fig.10.6
7) A beam of light is incident through the holes on side A and emerges out of the holes on the other face of the box as shown in the Figure 10.4. Which of the following could be inside the box?
(a) Concave lens
(b) Rectangular glass slab
(c) Prism
(d) Convex lens
Ans. (d) Convex lens
Since, when a parallel beam of ray is incident on a convex lens then all the emerging ray converges as shown in fig.10.7
Due to which convex lenses are also called as converging lenses. Hence, convex lens could be inside the box.
Fig.10.7
8) Which of the following statements is true?
(a) A convex lens has 4 dioptre power having a focal length 0.25 m
(b) A convex lens has –4 dioptre power having a focal length 0.25 m
(c) A concave lens has 4 dioptre power having a focal length 0.25 m
(d) A concave lens has –4 dioptre power having a focal length 0.25 m
Ans.(a) A convex lens has 4 dioptre power having a focal length 0.25 m Since, the power of lens P of focal length f is given by,
P = 1/f
If P = 4 dioptre then f in meters = ? (Since, power of convex lens is positive,)
f = 1/P
f = ¼
f = 0.25 m
9) Magnification produced by a rear view mirror fitted in vehicles
(a) Is less than one
(b) Is more than one?
(c) is equal to one
(d) Can be more than or less than one depending upon the position of the object in front of it
Ans. (a) Is less than one
Since, the rear view mirrors fitted in vehicles are convex mirrors.
Then, magnification m = Height of Image/ Height of object
m = h’/ h
m is –ve for real image and m is +ve for virtual image.
And the image produced by convex mirror is diminished, virtual and erect.
Here, the virtual image is formed and hence m is +ve.
h’ < h
m = h’ / h
m< 1
Thus, magnification is less than one.
10) Rays from Sun converge at a point 15 cm in front of a concave mirror. Where an object should be placed so that size of its image is equal to the size of the object?
(a) 15 cm in front of the mirror
(b) 30 cm in front of the mirror
(c) Between 15 cm and 30 cm in front of the mirror
(d) More than 30 cm in front of the mirror
Ans. (b) 30 cm in front of the mirror
In case of concave mirror, when the object is placed at center of curvature then the size of image produced will be equal to size of object as shown in fig.10.10
And we know that, focal point is the midpoint or center point of the radius of curvature of spherical mirror.
Given that, d (FP) = 15 cm
Hence, d (CP) = 2 (FP)
d (CP) = 30 cm
Hence to get the image of same size to that of object size, we have to put the object at a distance 30 cm in front of concave mirror.
Fig.10.10
11) A full length image of a distant tall building can definitely be seen by using
(a) A concave mirror
(b) A convex mirror
(c) A plane mirror
(d) Both concave as well as plane mirror
Ans. (b) A convex mirror
Since, the convex mirror produces the diminished, virtual and erect image.
So that, to see full length image of a distant tall building we can use a convex mirror only.
12) In torches, search lights and headlights of vehicles the bulb is placed
(a) Between the pole and the focus of the reflector
(b) Very near to the focus of the reflector
(c) Between the focus and center of curvature of the reflector
(d) At the center of curvature of the reflector
Ans. (b) Very near to the focus of the reflector Since, torches, search lights and headlights of vehicle produces a parallel beam of light due to which we can see the things in front of vehicles easily. The mirrors used are concave mirrors. And if we place the bulb very near to the focus of reflector then only we get the parallel beam of light which is highly concentrated. Since, when a ray of light is passed through focus of concave mirror the light rays parallel to principal axis are reflected as shown in fig.10.12
Fig.10.12
13) The laws of reflection hold well for
(a) Plane mirror only
(b) Concave mirror only
(c) Convex mirror only
(d) All mirrors irrespective of their shape
Ans.(d) All mirrors irrespective of their shape
Because, the laws of reflection does not depends on the shapes of the mirrors but depends on the optical properties of the mirrors only.
14) The path of a ray of light coming from air passing through a rectangular glass slab traced by four students are shown as A, B, C and D in Figure 10.5. Which one of them is correct?
(a) A
(b) B
(c) C
(d) D
Ans. (b) B
In fig. B, the incident ray travels from rarer (air) to denser (glass) medium so it bends towards the normal and when refracted ray travels from denser (glass) to rarer (air) medium it bends away from the normal.
Hence the correct answer is fig. B because it follows the laws of refraction.
15) You are given water, mustard oil, glycerin and kerosene. In which of these media a ray of light incident obliquely at same angle would bend the most?
(a) Kerosene
(b) Water
(c) Mustard oil
(d) Glycerin
Ans. (d) Glycerin
The extent of bending of ray of light depends on the refractive index of the medium i.e. indirectly on the optical density of the medium. The bending is more if refractive index is more. And here the refractive index of glycerin is more as compared to others.
16) Which of the following ray diagrams is correct for the ray of light incident on a concave mirror as shown in Figure 10.6?
(a) Fig. A
(b) Fig. B
(c) Fig. C
(d) Fig. D
Ans. (d) Fig. D
Because, in case of concave mirror when a ray of light parallel to principal axis is incident then the reflected ray must pass through the focus of the concave mirror which satisfies fig. D only.
Fig. D
17) Which of the following ray diagrams is correct for the ray of light incident on a lens shown in Fig. 10.7?
(a) Fig. A.
(b) Fig. B.
(c) Fig. C.
(d) Fig. D.
Ans.(a) Fig. A.
Since, in case of convex lens when the ray of light is passed through the focus of convex lens then emergent ray must be parallel to the principal axis of lens which satisfy fig. A only.
Fig. A
18) A child is standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body of the same size and that of the legs smaller. The following is the order of combinations for the magic mirror from the top.
(a) Plane, convex and concave
(b) Convex, concave and plane
(c) Concave, plane and convex
(d) Convex, plane and concave
Ans. (c) Concave, plane and convex
Since, concave mirrors are used to produce larger image & hence her head appears larger. The plane mirrors are used to produce image of same size as that of the object size & hence her middle portion of body will appears of same size. The convex mirrors are used to produce the image of small size than the object size and hence her legs appears smaller. Hence the correct answer is Concave, plane and convex.
19) In which of the following, the image of an object placed at infinity will be highly diminished and point sized?
(a) Concave mirror only
(b) Convex mirror only
(c) Convex lens only
(d) Concave mirror, convex mirror, concave lens and convex lens
Ans. (d) Concave mirror, convex mirror, concave lens and convex lens
Since, when the object is placed at infinity then the rays coming from infinity will be parallel to principal axis and after reflection or refraction through concave mirror, convex mirror, concave lens and convex lens passes through the focus and the image produced is highly diminished and point sized.
Short Answer Questions:
20) Identify the device used as a spherical mirror or lens in following cases, when the image formed is virtual and erect in each case.
(a) Object is placed between device and its focus, image formed is enlarged and behind it.
Ans: The device used is Concave mirror
Position of object: Between P & F
Position of the image: Behind the mirror
Size of the image: Enlarged
Nature of image: Virtual & erect
(b) Object is placed between the focus and device, image formed is enlarged and on the same side as that of the object.
Ans: The device used is convex lens.
Position of object: Between F1& optical center O
Position of the image: on the same side of the lens as the object
Size of the image: Enlarged
Nature of image: Virtual & erect
c) Object is placed between infinity and device, image formed is diminished and between focus and optical center on the same side as that of the object.
Ans: The device used is concave lens.
Position of object: Between infinity & optical center O
Position of the image: Between F1& O
Size of the image: Diminished
Nature of image: Virtual & erect
d) Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it.
Ans: The device used is convex mirror.
Position of object: Between infinity & pole P of the mirror
Position of the image: Between P & F behind the mirror
Size of the image: Diminished
Nature of image: Virtual & erect
21) Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram.
Ans:
Because, we know that, in case of refraction through rectangular glass slab, the extent of bending of ray at a point of incidence is equal to the extent of bending of ray at a point of emergence. Due to which the incident ray and emergent ray will be parallel to each other as shown in fig. This can be explained with help of diagram as given below.
22) A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent, if instead of water we use liquids like, kerosene or turpentine? Support your answer with reason.
Ans: When a pencil is immersed in water it appears bent due to the refraction of light at air-water interface. Since according to laws of refraction, when a ray of light travels from rarer i.e. air medium to denser i.e. water medium it bends towards the normal, due to this reason pencil bends at air-water interface.
But the extent of bending depends on the refractive index of the medium, which is different for different media. Hence, the extent of bending is indirectly depending on speed of light which is different for different media.
The refractive index of kerosene or turpentine is different and also the speed of light is different in both due to which the extent of bending is different for the pencil immersed in water and in kerosene or turpentine.
23) How is the refractive index of a medium related to the speed of light? Obtain an expression for refractive index of a medium with respect to another in terms of speed of light in these two media?
Ans:
The ray of light travelling from one transparent medium to second transparent medium, changes its path in second medium. The amount of change in path takes place during refraction for a given pair of media is expressed in terms of refractive index which is constant for a given medium.
The refractive index of the medium depends on the speed of light in that medium. The speed of light in vacuum is largest and which is 3* 108 m/s. In air it decreases slowly and more in water and glass.
Medium 1 (Air)
Medium 2 (Glass)
Let us consider that, the ray of light AB is incident on the air-glass interface & it refracted as BC. Let V1 and V2 be the speed of light in air & glass medium respectively.
Then the refractive index of second medium with respect to first is the ratio of speed of light in medium 1 to the speed of light in medium 2.
It is given by, μ21 = speed of light in medium 1/ speed of light in medium 2
= V1/V2
Similarly, the refractive index of medium 1 with respect to medium 2 is given by,
Μ12 = speed of light in medium 2/ speed of light in medium 1 = V2/V1
24) Refractive index of diamond with respect to glass is 1.6 and absolute refractive index of glass is 1.5. Find out the absolute refractive index of diamond.
Ans: The refractive index of diamond with respect to glass= refractive index of diamond / refractive index of glass
1.6 = refractive index of diamond/1.5
Refractive index of diamond = 1.6* 1.5 =2.4
Hence the absolute refractive index of diamond is 2.4
25) A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?
Ans: Focal length of the spherical mirror:
The distance between the principal focus and the pole of the mirror is called as the focal length of the spherical mirror.
It is denoted by f.
The statement is correct because convex lens having focal length 20cm produces magnified virtual & real image.
Position of object: Between F1 and C1 (2F1)
Position of the image: Beyond C2 (2F2)
Size of the image: Enlarged
Nature of image: Real & inverted
Hence, to obtain the magnified & real image the object should be placed between F1 and C1 i.e. at a distance between 20cm to 40cm.
Position of object: Between F1& optical center O
Position of the image: on the same side of the lens as the object
Size of the image: Enlarged
Nature of image: Virtual and erect
Hence, to obtain the magnified & virtual image the object should be placed between F1 and optical center O i.e. at a distance less than 20cm from the lens.
26) Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of the building? What is the approximate focal length of this lens?
Ans:
Position of object: At infinity
Position of the image: At focus F2
Size of the image: Highly diminished & point sized
Nature of image: Real & inverted
She must have to move the screen towards the optical center of the lens to obtain the sharp image of the building.
Because, window pane is beyond C2 which forms the image between F2 and C2 as shown in above fig.
To see the full building image which is outside the window i.e. at infinity, then the image will be obtained at focus F1, only if the screen is moved towards the optical center O.
The approximate focal length of convex lens will be 15cm.
27) How are power and focal length of a lens related? You are provided with two lenses of focal length 20 cm and 40 cm respectively. Which lens will you use to obtain more convergent light?
Ans: The power P of lens in terms of focal length f of lens is given by,
P = 1/ f
From above equation, the lens having higher power P i.e. having smaller focal length will produces more convergent light.
Hence, here the lens of focal length 20cm is used for producing more convergent light.
28) Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever may be angle of incidence. Show the same with the help of diagram.
Ans: When two plane mirrors are placed perpendicular to each other, then only the incident ray and reflected ray will always be parallel to each other, whatever may be angle of incidence which is explained in the figure given below.
29) Draw a ray diagram showing the path of rays of light when it enters with oblique incidence (i) from air into water; (ii) from water into air.
Ans:
Fig. a) shows the path of ray of light from air to water and fig.b) shows the path of ray of light from water to air.
Long Answer Questions
30) Draw ray diagrams showing the image formation by a concave mirror when an object is placed
Ans: (a) Between pole and focus of the mirror
Position of object: Between P & F
Position of the image: Behind the mirror
Size of the image: Enlarged
Nature of image: Virtual & erect
b) Between focus and center of curvature of the mirror
Position of object: Between C and F
Position of the image: Beyond C
Size of the image: Enlarged
Nature of image: Real & inverted
(c) At center of curvature of the mirror
Position of object: At C
Position of the image: At C
Size of the image: Same size
Nature of image: Real & inverted
(d) a little beyond center of curvature of the mirror
Position of object: Beyond C
Position of the image: Between F and C
Size of the image: Diminished
Nature of image: Real & inverted
(e) At infinity
Position of object: At infinity
Position of the image: At the focus
Size of the image: Highly diminished & point size
Nature of image: Real & inverted
31) Draw ray diagrams showing the image formation by a convex lens when an object is placed
Ans:
(a) Between optical center and focus of the lens
Position of object: Between F1& optical center O
Position of the image: on the same side of the lens as the object
Size of the image: Enlarged
Nature of image: Virtual and erect
(b) Between focus and twice the focal length of the lens
Position of object: Between F1 and C1 (2F1)
Position of the image: Beyond C2 (2F2)
Size of the image: Enlarged
Nature of image: Real & inverted
c) At twice the focal length of the lens
Position of object: At C1 (2F1)
Position of the image: At C2 (2F2)
Size of the image: Same size
Nature of image: Real & inverted
d) At infinity
Position of object: At infinity
Position of the image: At focus F2
Size of the image: Highly diminished & point sized
Nature of image: Real & inverted
e) At the focus of the lens
Position of object: At F1
Position of the image: At infinity
Size of the image: Highly enlarged
Nature of image: Real & inverted
32) Write laws of refraction. Explain the same with the help of ray diagram, when a ray of light passes through a rectangular glass slab.
Ans: The laws of refraction are as follows:
i) The incident ray, refracted ray and normal to the interface separating two media all lie in the same plane.
ii) The ratio of sine of angle of incidence to the sine of angle of refraction is constant and this constant is called as refractive index of the second medium with respect to first medium.
This law is also called as Snell’s law of refraction & it is valid for 0<i<90.
Then, sini/sinr = constant
i-Angle of incidence
r- Angle of refraction
sini/sinr= Refractive index of second medium with respect to first medium
The ray of light travelling from one transparent medium to second transparent medium, changes its path in second medium. The amount of change in path takes place during refraction for a given pair of media is expressed in terms of refractive index which is constant for a given medium.
Medium 1 (Air)
Medium 2 (Glass)
The above figure explains the laws of refraction when a ray of light is passed through the rectangular glass slab in detail.
33) Draw ray diagrams showing the image formation by a concave lens when an object is placed
Ans:
(a) At the focus of the lens
Position of object: At F1
Position of the image: Between F1& O
Size of the image: Diminished
Nature of image: Virtual & erect
(b) Between focus and twice the focal length of the lens
Position of object: Between F1 & C1 (2F1)
Position of the image: Between F1& O
Size of the image: Diminished
Nature of image: Virtual & erect
c) Beyond twice the focal length of the lens
Position of object: Beyond C1 (2F1)
Position of the image: At F1
Size of the image: Highly diminished & point sized
Nature of image: Virtual & erect
34) Draw ray diagrams showing the image formation by a convex mirror when an object is placed
Ans:
a) At infinity
Position of object: At infinity
Position of the image: At the focus F behind the mirror
Size of the image: Highly diminished & point size
Nature of image: Virtual & erect
b) At finite distance from the mirror
Position of object: Between infinity & pole P of the mirror
Position of the image: Between P & F behind the mirror
Size of the image: Diminished
Nature of image: Virtual & erect
35) The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?
Ans: Image distance v = 80 cm
Magnification m = -3
Object distance =?
Magnification of lens m = v/u = 80/u
-3 = 80/u
u = -26.66cm
Hence, the nature of the image will be real, magnified and inverted.
By lens formula,
1/v – 1/u = 1/f
1/80 – (-3/80) = 1/f
1/80 + 3/80 = 1/f
4/80 = 1/f
f = 20cm
The focal length is positive and hence the lens used is convex.
36) Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?
Ans: If the mirror used is concave then,
Focal length = -20cm
Magnification m = – 1/3
Also, m = -v/u
-1/3 = -v/u
V = u/3
By mirror formula,
1/v + 1/u = 1/f
3/u + 1/u = 1/-20
4/u = -1/20
u = -20 * 4
u = -80 cm
Hence the object should be place at a distance 80 cm from the concave mirror.
And the nature of the image is real, inverted and diminished.
If the mirror used is convex then,
Focal length = 20cm
Magnification m = 1/3
Also, m = -v/u
1/3 = -v/u
V =- u/3
By mirror formula,
1/v + 1/u = 1/f
-3/u + 1/u = 1/20 -2/u = 1/20
u = -20 * 2
u = -40 cm
Hence the object should be place at a distance 80 cm from the convex mirror.
And the nature of the image is virtual, erect and diminished.
37) Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of –50 cm. What is the nature of the lens and its power used by each of them?
Ans:
The ability of a lens to bend all the rays passing through it is known as power of lens.
Power of lens P is the reciprocal of focal length of the lens and its unit is diopter.
For Focal length = 50cm,
P = 100/f = 100/ 50 = 2 diopter
Hence lens is convex, since focal length of convex lens is positive.
For Focal length = -50cm,
P = 100/f = 100/ -50 = -2 diopter
Hence lens is concave, since focal length of concave lens is negative.
38) A student focused the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle screen and the lens as under
Position of candle = 12.0 cm
Position of convex lens = 50.0 cm
Position of the screen = 88.0 cm
Ans:
Candle (12cm)
Convex lens (50cm)Screen (88cm)
Distance between candle & lens = Object distance u
= 50-12
Object distance u = 38 cm
Distance between lens & screen = image distance v
= 88 – 50
Image distance v = 38 cm
By sign conventions, u = -38 cm and v= 38 cm
i) What is the focal length of the convex lens?
By lens formula,
1/v – 1/u = 1/f
1/38 + 1/38 = 1/f
f= 19 cm
Hence, the focal length of convex lens is 19cm.
ii) Where will the image be formed if he shifts the candle towards the lens at a position of 31.0 cm?
Candle (31cm)
Convex lens (50cm) Screen (88cm)
Distance between candle & lens = Object distance u
= 50-31
Object distance u = 19 cm
Distance between lens & screen = image distance v
= 88 – 50
Image distance v = 38 cm
By sign conventions, u = -19cm and v= 38 cm and f= 19 cm
Then by lens formula,1/v – 1/u = 1/f
1/v + 1/ 19 = 1/ 19
1/v = 0
V = infinity
Hence, the image will be formed at infinity.
iii) What will be the nature of the image formed if he further shifts the candle towards the lens?
If again the candle is shift towards lens i.e. it is placed in between focus and optical center of convex lens, then the image formed is magnified, virtual and erect.
iv) Draw a ray diagram to show the formation of the image in case (iii) as said above.
Position of object: Between F1& optical center O
Position of the image: on the same side of the lens as the object
Size of the image: Enlarged
Nature of image: Virtual and erect