## Light

## Light (O Level)

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Question 1 |

AX | |

BX | |

XC | |

AC |

Question 1 Explanation:

The normal is an imaginary line at the point of reflection and is perpendicular to the surface of reflection. It is BX.

Question 2 |

From the diagram below, given that PQ is a plane mirror, which is the incident ray and the angle of incidence?

Incident ray Angle of Incidence
AX AXP | |

AX AXB | |

XC BXC | |

XC CXQ |

Question 2 Explanation:

The light ray that strikes the mirror is known as the incident ray.
Angle of incidence is the angle between the normal and the incident ray.

Question 3 |

From the diagram below, given that PQ is a plane mirror, which is the reflected ray and the angle of reflection?

Reflected ray Angle of reflection
AX AXP | |

AX AXB | |

XC BXC | |

XC CXQ |

Question 3 Explanation:

The light ray that reflects away from the mirror is known as the reflected ray. Angle of reflection is the angle between the normal and the reflected ray.

Question 4 |

40 deg | |

50 deg | |

80 deg | |

90 deg |

Question 4 Explanation:

Angle of incidence is the angle between the normal and the incident ray. It is 50 deg

Question 5 |

20 deg | |

40 deg | |

70 deg | |

90 deg |

Question 5 Explanation:

Angle of incidence is the angle between the normal and the incident ray. It is 20 deg

Question 6 |

20 deg | |

40 deg | |

70 deg | |

90 deg |

Question 6 Explanation:

Angle between the incident ray and the reflected ray equals to double the angle of incidence or double the angle of reflection. It is 40 deg.

Question 7 |

A | |

B | |

C | |

D |

Question 7 Explanation:

For reflection, the angle of incidence equals to the angle of reflection. Choice A and Choice D both obey the law but choice D has the wrong incident ray in the wrong direction which does not show that it is a case of reflection.

Question 8 |

An incident ray strikes a plane mirror at an angle of incidence of 40

^{o}. What is the decrease in the angle of reflection if the incident ray moves to an angle of incidence of 30^{o}?10 deg | |

20 deg | |

30 deg | |

40 deg |

Question 8 Explanation:

For reflection, the angle of incidence equals to the angle of reflection. Therefore, when the angle of incidence decreases by 10 deg., the angle of reflection would also decrease by 10 deg.

Question 9 |

A plane mirror is inclined at 40

^{o}to the floor. An incident ray parallel to the floor strikes the mirror and a reflected ray is formed. What is the angle of reflection?20 deg | |

40 deg | |

50 deg | |

80 deg |

Question 9 Explanation:

Angle of reflection is the angle between the normal and the reflected ray.
Angle of incident = angle of reflection
= 50 deg

Question 10 |

A | |

B | |

C | |

D |

Question 10 Explanation:

The image seen through a plane mirror must be as far away (perpendicular to the plane of the mirror) inside the mirror as the object is (perpendicular to the plane of the mirror) outside the mirror.

Question 11 |

Four light bulbs are concealed from an observer by an opaque wall as shown. Without shifting the positions os the observer and the bulbs, how many bulbs can the observer see from the mirror?

1 | |

2 | |

3 | |

4 |

Question 11 Explanation:

By drawing two rays as shown, the images that are within the rays would be the images that the observer can see. There are altogether three images within the range.

Question 12 |

Which of the following are the properties of a plane mirror image?

- The image is the same size as the object.
- The image is virtual.
- The image is inverted.

1 only | |

1 and 2 only | |

1 and 3 only | |

1, 2 and 3 |

Question 12 Explanation:

The image is not inverted. The image is laterally inverted. The rest of the choices are the correct properties of a plane mirror image.

Question 13 |

Which of the following are the properties of a plane mirror image?

- The image is at the same distance as the object.
- The image is upright.
- The image is laterally inverted.

1 only | |

1 and 2 only | |

2 and 3 only | |

1, 2 and 3 |

Question 13 Explanation:

Choice 1 has basically no meaning. The phrase “The image is at the same distance as the object” does not give a reference point to measure a distance. It should be rewritten as “The perpendicular distance from the image to the plane of the mirror.”

Question 14 |

Which of the following is the mirror image of the word “EXAMPLE” when it is placed facing the plane mirror?

A | |

B | |

C | |

D |

Question 14 Explanation:

The image of a word as seen from a plane mirror can be obtained by first writing the word on a piece of thin paper than look at the word from the opposite side of the thin paper.

Question 15 |

5 : 20 | |

6 : 40 | |

11 : 50 | |

12 : 10 |

Question 16 |

Two men are standing in front of a plane mirror as shown. When a man looks into the mirror, how far away from him will man B seem to be?

12 m | |

16 m | |

26 m | |

28 m |

Question 16 Explanation:

The distance from A to man B is 16 m.

Question 17 |

Two men are standing in front of a plane mirror as shown. If man A walks 5 m backward and then looks into the mirror, how far away from him will man B seem to be?

7 m | |

16 m | |

21 m | |

26 m |

Question 17 Explanation:

The distance from A to man B is 21 m.

Question 18 |

Two men are standing in front of a plane mirror as shown. If man B walks 5 m backward and then looks into the mirror, how far away from him will man B seem to be?

9 m | |

11 m | |

18 m | |

23 m |

Question 18 Explanation:

The distance from A to man B is 11 m.

Question 19 |

Man A is standing in front of a plane mirror while man B is running towards him from behind. If man B is running at a speed of 1 m/s, how many meters nearer does man B seem to be away from man A after 5 seconds?

1 m | |

5 m | |

6 m | |

10 m |

Question 19 Explanation:

When B runs for 5 seconds at a rate of 1 m/s, he has moved 5 m nearer to man A. One characteristic of the image formed by a plane mirror is: the perpendicular distance from the object to the plane of the mirror is the same as the perpendicular distance from the image to the plane of the mirror. Therefore, the observer would see man B to be 5 m nearer.

Question 20 |

Man A is facing a plane mirror while man B is running towards him from behind. If man B is running at a speed of 2 m/s, how fast does man B seem to be running towards man A?

1 m/s | |

2 m/s | |

3 m/s | |

4 m/s |

Question 20 Explanation:

One characteristic of the image formed by a plane mirror is that the perpendicular distance from the object to the plane of the mirror is the same as the perpendicular distance from the image to the plane of the mirror. When man B runs at a rate of 2 m/s towards A, his image would also be seen running out of the mirror at 2 m/s every second.

Question 21 |

A man is running towards a plane mirror at a speed of 2 m/s. How fast does he see himself running towards his image?

1 m/s | |

2 m/s | |

3 m/s | |

4 m/s |

Question 21 Explanation:

One characteristic of the image formed by a plane mirror is that the perpendicular distance from the object to the plane of the mirror is the same as the perpendicular distance from the image to the plane of the mirror. Assuming the distance between the man and his image would have each move 2 m towards the mirror. Effectively the distance between them is decreasing by 4 m/s. The speed of the man approaching his image is therefore 4 m/s.

Question 22 |

A man is standing still while a plane mirror is moving away from him at a speed of 4 m/s, how fast does he see his image moving away?

1 m/s | |

2 m/s | |

4 m/s | |

8 m/s |

Question 22 Explanation:

Assuming the distance between the man to the mirror at time t=0 s is x m. The distance between the image to mirror at time t=0 s is therefore also x m. After 1 s, the mirror would have moved 4 m away from the man. The image distance from the mirror would have also increased by 4 m. The distance between the man and his image therefore increases by 8 m in 1 s. The speed of the man moving away from his image is therefore 8 m/s.

Question 23 |

A man drills a tiny hole at the 40 cm mark of a meter long ruler. He places a 30 cm long plane mirror in front of the ruler as shown. What are the minimum and maximum readings he can read from the image of the ruler if he peeps through the tiny hole looking into the plane mirror?

Minimum Maximum
reading reading
20 cm 70 cm | |

20 cm 80 cm | |

30 cm 60 cm | |

30 cm 80 cm |

Question 23 Explanation:

In order for the eye to see the maximum reading and the minimum reading, the eye must look at the two extreme ends of the 30 cm mirror. The diagram below makes use of the fact that angle of incidence = angle of reflection to show the range that can be seen by the eye through the mirror. Take note that if the observer and the object is at the same perpendicular distance away from the mirror, the image seen through the mirror is twice as much as the size of the mirror. For this case a 30 cm mirror can see a range of 60 cm.

Question 24 |

A painter leans back against a pointed wall while looking into a 1 m long mirror at the opposite end of a rectangular room. How much of the painted wall can he see through the 1 m long mirror?

1 m | |

2 m | |

6 m | |

12 m |

Question 24 Explanation:

Assume that the painter is facing the centre of the mirror. To see the maximum range, he has to look at the two extreme ends of the mirror. By applying angle of incidence = angle of reflection, the maximum length of painted wall is 2 m.

Question 25 |

A painter standing at the centre of a rectangular room looking into a 1 m long mirror at the opposite end of the room. How much of the painted wall can he see through the 1 m long mirror?

1 m | |

2 m | |

3 m | |

6 m |

Question 25 Explanation:

Assume that the painter is facing the centre of the mirror. To see the maximum range, he has to look at the two extreme ends of the mirror. By applying angle of incidence = angle of reflection.
DC = 0.5 m => BD = 0.5 m (ABC is an isosceles triangle)
Triangle ADB is similar to triangle AFE,
AD/BD = AF/FE => 6/0.5 = 12/x
x = 1 m.
GH = EF = 1 m
The maximum length of painted wall is 1m + 1m + 1m = 3 m.

Question 26 |

A plane mirror AB is positioned at the corner of a road as shown in the plan view below. Which men can the observer see through mirror?

P and Q only | |

P and R only | |

Q and R only | |

P, Q and R |

Question 26 Explanation:

Draw the image of P, Q and R as P1, Q1 and R1 by using the extension of the mirror as the line of symmetry.
Draw two lines from the observer touching the two extreme ends of the mirror and then extend out further to show the range that the observer can see through mirror.
Only the images of Q and R are within the range.

Question 27 |

Two mirrors (facing each other) and an object O are places in the grid as shown. At which position(s) can the virtual image of object O be formed?

C only | |

D only | |

A and B only | |

B and C only |

Question 27 Explanation:

First draw the extension of mirror X and mirror Y.
Object O is facing mirror X and therefore based on the extension of mirror X image 1 Is formed at C. Image 1 is facing mirror Y and therefore based on the extension of mirror Y, image 2 is formed at B. Image 2 will further form image 3 based on mirror X and it will continue forever on and on. But for the question, only B and C can be the image positions of object O.

Question 28 |

A mirror periscope is used to observe a bird as shown below. How far away will the bird seem to be from the observer?

320 cm | |

470 cm | |

620 cm | |

940 cm |

Question 28 Explanation:

The distance travelled by the light ray is the distance the bird seems to be away from the observer.
Distance = 300 cm + 150 cm + 20 cm
= 470 cm

Question 29 |

Which are the incident ray, the refracted ray and the emerging ray?
Incident Refracted Emerging
Ray ray ray

Incident Refracted Emerging
Ray ray ray
AB BC CD | |

AB CD BC | |

CD BC AB | |

CD AB BC |

Question 29 Explanation:

The light ray that strikes the boundary of two mediums is known as the incident ray. The light ray that refracts away from the boundary of two mediums is known as the refracted ray. The light ray that comes out from a medium and refracts back into the original medium and refracts back into the original medium is known as the emerging ray.

Question 30 |

Incident Refracted
Ray ray
AB only BC only | |

AB and BC only CD only | |

AB only BC and CD only | |

AB and BC only BC and CD only |

Question 30 Explanation:

If AB is the incident ray, BC is known as the refracted ray. If BC is the incident rays and BC and CD can be refracted rays.

Question 31 |

Angle of Angle of
Incidence refraction
∠ABD ∠CBG | |

∠ABF ∠CBE | |

∠ABD ∠CBE | |

∠ABF ∠CBG |

Question 31 Explanation:

Angle of incidence is the angle between the normal and the incident ray.
Angle of refraction is the angle between the normal and the refracted ray.

Question 32 |

A ray of light travels from air to glass as shown below. Given that the refractive index of air is 1.0 and the refractive index of glass is 1.5, what is the angle of refraction,

*r*?22.6 deg | |

30.8 deg | |

35.3 deg | |

40.0 deg |

Question 32 Explanation:

n1 sin i = n2 sin r
(where n1 is the refractive index of the incident medium, n2 is the refractive index of the refracted medium, ‘i’ is the angle of incidence and ‘r’ is the angle of refraction)
(1)sin 60 = (1.5)sin(r)
sin r = 0.577
r = 35.3 deg

Question 33 |

A ray of light travels from air to water as shown below. Given that the refractive index of air is 1.0, what is the refractive index of water?

1.2 | |

1.3 | |

1.4 | |

1.5 |

Question 33 Explanation:

n1 sin i = n2 sin r
(where n1 is the refractive index of the incident medium, n2 is the refractive index of the refracted medium, ‘i’ is the angle of incidence and ‘r’ is the angle of refraction)
(1)sin 30 = (n2)sin(22.6)
n2 = 1.30

Question 34 |

A ray of light travels from air to water as shown below. Given that the refractive index of air is 1.0 and the refractive index of water is 1.3, what is the angle of incidence,

*i*?22.7 deg | |

23.1 deg | |

39.0 deg | |

40.5 deg |

Question 34 Explanation:

n1 sin i = n2 sin r
(where n1 is the refractive index of the incident medium, n2 is the refractive index of the refracted medium, ‘i’ is the angle of incidence and ‘r’ is the angle of refraction)
(1)sin i = (1.3)sin(30)
i = 40.5 deg

Question 35 |

A ray of light travels from water to air as shown below. Given that the refractive index of air is 1.0 and the refractive index of water is 1.3, what is the angle of incidence,

*i*?+22.6 deg | |

23.1 deg | |

39.0 deg | |

40.5 deg |

Question 35 Explanation:

n1 sin i = n2 sin r
(where n1 is the refractive index of the incident medium, n2 is the refractive index of the refracted medium, ‘i’ is the angle of incidence and ‘r’ is the angle of refraction)
(1.3)sin i = (1)sin(30)
i = 22.6 deg

Question 36 |

A ray of light travels from water to air as shown below. Given that the refractive index of air is 1.0 and the refractive index of water is 1.3, what is the angle of refraction,

*r*?22.6 deg | |

23.1 deg | |

39.0 deg | |

40.5 deg |

Question 36 Explanation:

n1 sin i = n2 sin r
(where n1 is the refractive index of the incident medium, n2 is the refractive index of the refracted medium, ‘i’ is the angle of incidence and ‘r’ is the angle of refraction)
(1.3)sin 30 = (1)sin(r)
r = 40.5 deg

Question 37 |

A ray of light travels from water to air as shown below. Given that the refractive index of air is 1.0, what is the refractive index of glass?

1.2 | |

1.3 | |

1.4 | |

1.5 |

Question 37 Explanation:

n1 sin i = n2 sin r
(where n1 is the refractive index of the incident medium, n2 is the refractive index of the refracted medium, ‘i’ is the angle of incidence and ‘r’ is the angle of refraction)
(n1)sin 40 = (1)sin(74.6)
n1 = 1.5

Question 38 |

A ray of light travels from water to glass as shown below. Given that the refractive index of water is 1.3 and the refractive index of glass is 1.5, what is the angle of refraction?

30.7 deg | |

35.3 deg | |

41.7 deg | |

48.6 deg |

Question 38 Explanation:

n1 sin i = n2 sin r
(1.3)sin 60 = (1.5)sin(r)
sin r = 0.577
r = 48.6 deg

Question 39 |

A ray of light travels from water to diamond as shown below. Given that the refractive index of water is 1.3, what is the refractive index of diamond?

1.5 | |

1.7 | |

2.2 | |

2.3 |

Question 39 Explanation:

n1 sin i = n2 sin r
(1.3)sin 30 = (n2)sin(17.2)
n2 = 2.2 deg

Question 40 |

A ray of light travels from diamond to glass as shown below. Given that the refractive index of diamond is 2.0 and the refractive index of glass is 1.5, what is the angle of incidence?

14.5 deg | |

19.5 deg | |

22.0 deg | |

22.5 deg |

Question 40 Explanation:

n1 sin i = n2 sin r
(2.0)sin i = (1.5)sin(30)
i = 22.0 deg

Question 41 |

Ray AB Ray AC
Reflected ray Refracted ray | |

Reflected ray Reflected ray | |

Refracted ray Reflected ray | |

Refracted ray Refracted ray |

Question 41 Explanation:

When light ray travels from one medium to another medium of different refractive index, there will be a weak reflected ray (ray AB) at the boundary of the two mediums and the light ray that passes through will be refracted (ray AC).

Question 42 |

A ray of light travels from air to diamond as shown below. Given that the refractive index of air is 1.0 and the refractive index of diamond is 2.2, what are the ∠ABD and ∠EBC?

∠ABD ∠EBC
17 deg 17 deg | |

17 deg 40 deg | |

40 deg 17 deg | |

40 deg 40 deg |

Question 42 Explanation:

Angle of reflection = 40 deg
n1 sin i = n2 sin r
(1.0)sin 40 = (2.2)sin(r)
r = 17.0 deg

Question 43 |

A light ray travels from a medium P to medium Q. Given that medium P has a refractive index of 1.0 and medium Q has a refractive index of 1.2, which of the following is a possible refracted ray?

A | |

B | |

C | |

D |

Question 43 Explanation:

When a light ray travels from an optically less dense medium to an optically denser medium, the refracted ray will bend towards the normal.

Question 44 |

A light ray travels from a medium P to medium Q. Given that medium P has a refractive index of 1.5 and medium Q has a refractive index of 1.0, which of the following is/are possible refracted ray(s)?

2 only | |

1 and 2 only | |

2 and 3 only | |

3 and 4 only |

Question 44 Explanation:

When a light ray travels from an optically denser medium to an optically less dense medium, the refracted ray will bend away from the normal. When the angle of incidence increases till it reaches critical angle, the angle of refraction will be 90 deg. When angle of refraction = 90 deg, the refracted ray will travel along the boundary of the two mediums.

Question 45 |

A light ray travels from a medium P to medium Q. Given that medium P has a refractive index of 1.5 and medium Q has a refractive index of 1.2, which of the following rays is/are possible outcome(s)?

3 only | |

1 and 3 only | |

2 and 3 only | |

1, 2 and 3 only |

Question 45 Explanation:

When a light ray travels from an optically denser medium to an optically less dense medium, the refracted ray will bend away from the normal. When the angle of incidence increases till it reaches critical angle, the angle of refraction will be 90 deg. When angle of refraction = 90 deg, the refracted ray will travel along the boundary of the two mediums.
When the angle of incidence increases beyond the critical angle, total internal reflection will occur and all light rays will be reflected back into medium P with the angle of reflection = angle of incidence.

Question 46 |

Given that the refractive index of water is 1.0 and the refractive index of water is 1.0, what is the critical angle when a light ray travels from glass to air?

22.6 deg | |

30.0 deg | |

41.8 deg | |

48.6 deg |

Question 46 Explanation:

n1 sin i = n2 sin r
(1.5)sin (c) = (1.0)sin(90)
c = 41.8 deg

Question 47 |

Given that the critical angle angle of the medium is 45

^{o}, what is the refractive index of the medium?0.71 | |

1.00 | |

1.33 | |

1.41 |

Question 47 Explanation:

n1 sin i = n2 sin r
(n1)sin (45) = (1.0)sin(90)
n1 = 1.41

Question 48 |

A light ray travels from a medium of refractive index 1.5 to another medium of refractive index of 1.2. At what angle will total internal reflection occur?

22.6 deg | |

41.8 deg | |

53.1 deg | |

56.4 deg |

Question 48 Explanation:

n1 sin i = n2 sin r
(1.5)sin (c) = (1.2)sin(90)
c = 53.1 deg

Question 49 |

A light ray travels from a medium of refractive index 1.4 to another medium of refractive index of 2.2. At what angle will total internal reflection occur?

39.5 deg | |

41.8 deg | |

53.1 deg | |

Critical angle cannot be reached |

Question 49 Explanation:

When a light ray travels from an optically less dense medium to all denser mediums, it is impossible to achieve critical angle. The angle of incidence will reach 90 deg before the angle of refraction reaches 90 deg.

Question 50 |

A scientist is given 3 mediums to do a certain experiment. The refractive indices of the three mediums are 1.0, 1.2 and 1.4. What are the possible critical angles obtained by these three mediums?
1) 45.6

^{o}2) 56.4^{o}3) 59.0^{o}1 only | |

1 and 3 only | |

2 and 3 only | |

1, 2 and 3 |

Question 50 Explanation:

n1 sin i = n2 sin r
(1.2)sin (c) = (1.0)sin(90)
c = 56.4 deg
n1 sin i = n2 sin r
(1.4)sin (c) = (1.2)sin(90)
c = 59.0 deg
n1 sin i = n2 sin r
(1.4)sin (c) = (1.0)sin(90)
c = 45.6 deg

Question 51 |

A light ray travels through three mediums as shown. Which of the following shows the ascending order of refractive index?

Medium 1, Medium 2, Medium 3 | |

Medium 2, Medium 3, Medium 1 | |

Medium 3, Medium 1, Medium 2 | |

Medium 1, Medium 3, Medium 2 |

Question 51 Explanation:

When a light ray travels from medium A to medium B, the more the light ray is being refracted away from the normal, the lesser is the refractive index of the medium B.
By comparing the refraction os light ray in the 3 mediums, the light ray passing through medium 3 is refracted away the most from the normal and therefore it has the smallest refractive index. The refracted light ray that bends towards the normal in Medium 2 (indicating that medium 2 has a higher refractive index than medium 1), and therefore it has the largest refractive index.

Question 52 |

A light ray travels from medium 1 to 3 as shown. Which of the following is ascending order of refractive index?

Medium 1, Medium 2, Medium 3 | |

Medium 1, Medium 3, Medium 2 | |

Medium 3, Medium 1, Medium 2 | |

Medium 3, Medium 2, Medium 1 |

Question 52 Explanation:

Total internal reflection can only happen when light ray travels from a denser medium to a less dense medium. Medium 3 is therefore less dense than medium 2. Light ray will bend away from normal when light ray travels from a denser medium to a less dense medium. Medium 2 is therefore less dense than medium 1.

Question 53 |

1) A periscope uses two 45

^{o}prisms to reflect the light ray into the observer’s eye. What is the possible refractive index of the prism material? 1) 1.2 2) 1.5 3) 2.21 only | |

2 only | |

2 and 3 only | |

1, 2 and 3 |

Question 53 Explanation:

sin c = 1/n
sin c = 1/1.2
c = 56.4 deg
sin c = 1/n
sin c = 1/1.5
c = 41.8 deg
sin c = 1/n
sin c = 1/2.2
c = 27.0
The medium that can cause a total internal reflection at an angle of incidence of 45 deg. is the medium that is suitable to be used as the prism. The only medium that cannot achieve total internal reflection at i = 45 deg. is the one with refractive index of 1.2.

Question 54 |

A | |

B | |

C | |

D |

Question 54 Explanation:

Incident ray refracts towards the normal when it travels from air to glass, and refracts away from the normal when it travels from glass to air.

Question 55 |

A | |

B | |

C | |

D |

Question 55 Explanation:

Incident ray refracts towards the normal when it travels from air to glass, and refracts away from the normal when it travels from glass to air.

Question 56 |

Which of the following is the emergent ray when the incident ray strikes the rectangular glass block from air?

A | |

B | |

C | |

D |

Question 56 Explanation:

Incident ray refracts towards the normal when it travels from air to glass, and refracts away from the normal when it travels from glass to air. The emergent ray is parallel to the incident ray.

Question 57 |

A | |

B | |

C | |

D |

Question 57 Explanation:

Incident ray refracts towards the normal when it travels from air to glass, and refracts away from the normal when it travels from glass to air.

Question 58 |

A | |

B | |

C | |

D |

Question 58 Explanation:

Question 59 |

What is/are the effect(s) of using a thin converging lens instead of a thick converging lens?

- The image formed will be brighter.
- The focal length will be longer.
- The image formed will always be virtual.

2 only | |

1 and 2 only | |

2 and 3 only | |

1 and 3 only |

Question 59 Explanation:

1. The image will be brighter. (Incorrect)
Brightness of the image depends on the amount of light that passes through the lens and how focused the light is. Changing the thickness of the glass cannot change the brightness of the image.
2. The focal length will be longer. (Correct)
The thinner glass material is not able to refract the light ray as much as what a thicker lens can do. The emergent light rays therefore will travel further before they converge to a point.
3. The image formed will always be virtual. (Incorrect)
The thin converging lens can produce both real images and virtual images.

Question 60 |

Two light rays parallel to the principal axis strike a converging lens and converge to point B as shown. Which point is the focal point?

A | |

B | |

C | |

D |

Question 60 Explanation:

Focal point of a lens is defined as the point on the principal axis to which rays parallel and close to the principal axis converge after passing through the lens. Point B is a point on principal axis where the light rays converge

Question 61 |

The focal length is

the distance from the light source to the image. | |

the distance from the object to the centre of the lens. | |

the distance from the image to the centre of the lens. | |

the distance from the focal point to the centre of the lens. |

Question 61 Explanation:

Focal length is the distance from the focal point to the centre of the lens.

Question 62 |

Which of the following are true?

- The principal focus of a converging lens is real.
- The principal focus of a diverging lens is virtual.
- The principal focus of a lens is along the principal axis.

1 and 2 only | |

2 and 3 only | |

1 and 3 only | |

1, 2 and 3 |

Question 62 Explanation:

Facts. All three choices are true.

Question 63 |

Which of the following are true?

The image distance obtained from a far away object is the focal length of the convex lens. | |

The images obtained by a convex lens always lie on the focal point. | |

The images obtained by a convex lens are always real. | |

The images obtained by a convex lens are always inverted. |

Question 63 Explanation:

Focal length is the distance from the focal point to the centre of the lens. This distance can be determined approx. by placing the convex lens such that the rays coming from a far away object such as the sun fall on the lens. Adjust the distance from the lens to the floor to obtain a focused bright spot. This distance from the lens to the floor is the approximate focal length of the lens. The characteristics of image obtained using a convex lens depends on the object distance (the distance from the object to the lens). As the object distance changes, the image distance changes correspondingly.

Question 64 |

Which of the following are true?

- An incident ray parallel to the principal axis emerging from the converging lens will always pass through the focal point F of the lens.
- An incident ray that passes through the focal point of a converging lens will travel in parallel to the principal axis after emerging from converging lens.
- A ray that passes through the optical centre of a thin converging lens will not be refracted by the lens and continues to travel along its original path

1 and 2 only | |

2 and 3 only | |

1 and 3 only | |

1, 2 and 3 |

Question 64 Explanation:

Facts. All three choices are true.

Question 65 |

real, inverted and diminished | |

real, inverted and enlarged | |

virtual, inverted and diminished | |

virtual, upright and enlarged |

Question 65 Explanation:

When u < f, image is always virtual, upright and enlarged (magnified)

Question 66 |

real, inverted and diminished | |

real, inverted and enlarged | |

virtual, upright and enlarged | |

no image formed |

Question 66 Explanation:

When u = f, there is no image formed as the light rays do not converged of diverged.

Question 67 |

real, inverted and diminished | |

real, inverted and enlarged | |

virtual, upright and enlarged | |

no image formed |

Question 67 Explanation:

When f < u < 2f, image is always real, inverted and enlarged (magnified).

Question 68 |

real, inverted and diminished | |

real, inverted and same size as object | |

real, inverted and enlarged | |

virtual, upright and enlarged |

Question 68 Explanation:

When u = 2f, image is always real, inverted and as same size as the object.

Question 69 |

real, inverted and diminished | |

real, inverted and same size as object | |

real, inverted and enlarged | |

virtual, upright and enlarged |

Question 69 Explanation:

When u > 2f, image is always real, inverted and diminished.

Question 70 |

Which of the following are always true?

- When the object is placed nearer to the focal point, the image gets bigger.
- When the object is placed further away from the converging lens, the image gets smaller.
- When the object is placed further away from the converging lens, the image distance approaches focal length.

1 and 2 only | |

2 and 3 only | |

1 and 3 only | |

1, 2 and 3 |

Question 70 Explanation:

1. When the object is placed to the focal point, the image gets bigger. (Correct)
When the object approaches the focal point from the converging lens, the image becomes a bigger virtual image. When the object approaches the focal point from infinity, the image becomes a bigger real image.
2. When the object is placed further away from the converging lens, the image gets smaller. (Incorrect)
This is only partially true. When the object is placed between the converging lens and the focal point and as the object is moving away from the converging lens, the object is actually moving nearer to the focal point and the virtual image will get better.
3. When the object is placed very far away from the converging lens, the image distance approaches focal length. (Correct)
As the object is moving away from the focal point towards infinity, the real image is getting from big to small and the object distance also reduces from infinity to a minimum length of one focal length.

Question 71 |

A | |

B | |

C | |

D |

Question 71 Explanation:

Real images can only form when there is light ray converging on it. From the diagram below, the only possible object to form the image I is object D. Light ray that passes will continue to move straight without being refracted.

Question 72 |

A | |

B | |

C | |

D |

Question 72 Explanation:

Real images can only form when there is light ray converging on it. From the diagram below, the only possible object to form the image I is lens B. Light ray that passes the optical centre of a thin converging lens will continue to move straight without being refracted.

Question 73 |

A | |

B | |

C | |

D |

Question 73 Explanation:

Real images can only form when there is light ray converging on it. From the diagram below, the only possible object to form the image I is lens C. Light ray that passes the optical centre of a thin converging lens will continue to move straight without being refracted.

Question 74 |

A light ray parallel to the principal axis travels into a converging lens as shown. Which is the emergent ray from the converging lens?

A | |

B | |

C | |

D |

Question 74 Explanation:

An incident ray parallel to the principal axis after emerging from the converging lens will always pass through the focal point F of the lens.

Question 75 |

A light ray travels into a converging lens as shown. Which is the emergent ray from the converging lens?

A | |

B | |

C | |

D |

Question 75 Explanation:

An incident ray that passes through the focal point of a converging lens will travel in parallel to the principal axis after emerging from the converging lens.

Question 76 |

A light ray travels into a converging lens as shown. Which is the emergent ray from the converging lens?

A | |

B | |

C | |

D |

Question 76 Explanation:

An incident ray parallel to the principal axis after emerging from the converging lens will always pass through the focal point F of the lens.
An incident ray that passes through the focal point of a converging lens will travel in parallel to the principal axis after emerging from the converging lens.
Now that the angle of incidence is in between the two cases mentioned above, the emergent ray should be between choice B and choice D.

Question 77 |

A light ray travels into a converging lens as shown. Which is the emergent ray from the converging lens?

A | |

B | |

C | |

D |

Question 77 Explanation:

An incident ray that passes through the focal point of a converging lens will travel in parallel to the principal axis after emerging from the converging lens.
Now that the angle of incidence is bigger than that, the emergent ray should be in a choice that shows a greater angle from the incident ray than choice B.

Question 78 |

A converging lens has a focal length of 15 cm. Given that an object is placed 10 cm from the optical centre of the lens, what are the characteristics of the image formed?

real, inverted and diminished | |

real, inverted and same size as object | |

real, inverted and enlarged | |

virtual, upright and enlarged |

Question 78 Explanation:

Focal length f is 15 cm.
Object distance u is 10 cm.
For u < f, the image will be virtual, upright and magnified.

Question 79 |

A converging lens has a focal length of 15 cm. Given that an object is placed 31 cm from the optical centre of the lens, what are the characteristics of the image formed?

real, inverted and diminished | |

real, inverted and same size as object | |

real, inverted and enlarged | |

virtual, upright and enlarged |

Question 79 Explanation:

Focal length f is 15 cm.
Object distance u is 31 cm.
For u < 2f, the image will be real, inverted and diminished.

Question 80 |

A converging lens has a focal length of 15 cm. Given that an object is placed 5 m from the optical centre of the lens, what are the characteristics of the image formed?

real, inverted and diminished | |

real, inverted and same size as object | |

real, inverted and enlarged | |

virtual, upright and enlarged |

Question 80 Explanation:

Focal length f is 15 cm.
Object distance u is 5 m.
For u >> 2f, the image will also be real, inverted and diminished. When the object distance is greater than 20f, it can be considered as u >> 2f. The special thing here is that when we have about u >> 2f is the image is being formed is almost at the focal point.

Question 81 |

A converging lens has a focal length of 15 cm. Given that an object is placed 10 cm from the optical centre of the lens, what is the image distance (the distance from the image to the optical centre)?

10 cm | |

20 cm | |

40 cm | |

cannot be determined |

Question 81 Explanation:

Focal length f is 20 cm. Object distance u is 10 m.
For u >> 2f, the image will also be real, inverted and diminished. When the object distance is greater than 20f, it can be considered as u >> 2f. What is special here is that when we have u >> 2f, the image formed is almost at the focal point.
The special thing about u >> 2f is that the image is formed almost at the focal point.

Question 82 |

When an object is placed 29 cm from the optical centre of the converging lens, the image formed is real, inverted and magnified but when the object is placed 31 cm from the optical centre of the lens, the image formed is real, inverted and diminished. What is the focal length of this converging lens?

15 cm | |

30 cm | |

60 cm | |

cannot be determined |

Question 82 Explanation:

For the image to be real, inverted and magnified, the object distance f < u < 2f (2f > 29 cm). For the image to be real, inverted and diminished, the object distance u > 2f (2f < 31 cm).
It therefore can be concluded that 2f = 30 cm. The focal length f = 15 cm.

Question 83 |

When an object is placed 11 cm from the optical centre of the converging lens, the image formed is real, inverted and magnified but when the object is placed 9 cm from the optical centre of the lens, the image formed is virtual, upright and magnified. What is the focal length of this converging lens?

5 cm | |

10 cm | |

20 cm | |

cannot be determined |

Question 83 Explanation:

For the image to be real, inverted and magnified, the object distance f < u < 2f (f > 11 cm).
For the image to be virtual, upright and magnified, the object distance u < f (f > 9 cm). It therefore can be concluded that f = 10 cm. The focal length f = 10 cm.

Question 84 |

An overhead projector in the classroom uses a converging lens to produce the image on the screen. If the focal length of the lens is 10 cm, where should a transparency be placed from the converging lens to form a clear image?

5 cm | |

10 cm | |

15 cm | |

20 cm |

Question 84 Explanation:

The image obtained from an overhead projector is real, inverted and magnified. To achieve such image, it must be a case of f < u < 2f. This implies that 10 cm < u < 20 cm.

Question 85 |

A camera uses a converging lens to produce an image on the film. If the focal length of the lens is 10 cm, where should an object be placed from the camera for photo taking?

10 cm | |

15 cm | |

20 cm | |

100 cm |

Question 85 Explanation:

The image obtained from a camera is real, inverted and diminished. To achieve such image, it must be a case of u > 2f. This implies that u > 20 cm.

Question 86 |

A photocopy machine uses a converging lens to focus images for printing on papers. If the focal length of the lens is 6 cm, what is the distance from the original copy to the lens during photocopying?

2 cm | |

6 cm | |

10 cm | |

12 cm |

Question 86 Explanation:

To reproduce a similar print, the real image must be the same size as the object. It can only be achieved when u = 2f. This implies that u = 12 cm.

Question 87 |

A telescope uses a converging lens to see birds which are far away. If the focal length of the lens is 20 cm, what is the distance from the observer’s eye to the lens to see a clear image of the bird?

10 cm | |

20 cm | |

40 cm | |

500 cm |

Question 87 Explanation:

When the object is far away from the converging lens, the image distance will be at one focal length from the lens. The observer will need to place the eye at a distance of 1 focal length f (20 cm) away from the lens for the image to cast on his retina.

Question 88 |

An object O is placed near a mirror as shown. Where is the image of object O as seen by the observer?

A | |

B | |

C | |

D |

Question 88 Explanation:

The image of the object is independent of the observer. The image is always directly opposite to the object and its perpendicular distance from the mirror is the same as the perpendicular distance from the object to the mirror.

Question 89 |

The diagram below shows a ray of light being reflected from a plane mirror. Which of the labeled angles are the angle of incidence and the angle of reflection?

Angle of Angle of
Incidence reflection
1 2 | |

2 3 | |

1 4 | |

3 1 |

Question 89 Explanation:

Angle of incidence is defined as the angle between normal and the incident ray. Angle of reflection is defined as the angle between the normal and the reflected ray.
The angle between the normal and the incident ray is angle 2 and the angle between the normal and the reflected ray is angle 3.

Question 90 |

Which of the following is a virtual image?

- The image formed by an image projector.
- The image formed by a mirror.
- The upright image formed by a magnifying glass.

2 only | |

1 and 2 only | |

2 and 3 only | |

1, 2 and 3 |

Question 90 Explanation:

Virtual images are images that cannot be formed on the screen. This type of image is formed because the brain is being “tricked” by the light rays entering the eyes.
Real images are formed by light rays landing on a screen but virtual images are images that are formed by the perception of the mind and therefore there is no light ray landing at the palace that it seems to be.
Out of the 3 choices, only image projector emits light ray on a screen and formed real images.

Question 91 |

A plane mirror is inclined at 40

^{o}to the floor. An incident ray parallel to the floor strikes the mirror and a reflected ray is formed. If the angle of inclination is increased to 50^{o}without changing the direction of the ray, what is the change in angle between the incident ray and the reflected ray?5 deg | |

10 deg | |

20 deg | |

40 deg |

Question 91 Explanation:

Angle of reflection is the angle between the normal and the reflected ray.
When the inclination is 40 deg,
Angle of incident incidence = angle of reflection = 50 deg.
Angle between incident ray and reflected ray = 2 x 50 = 100 deg.
When the inclination is 50 deg,
Angle of incident incidence = angle of reflection = 40 deg.
Angle between incident ray and reflected ray = 2 x 40 = 80 deg.
The change in the angles between incident ray and reflected ray = 100 – 80 = 20 deg.

Question 92 |

The ray from a laser pointer hits a plane mirror and the reflected ray strikes a screen. How far away from the point X should the ray strike the mirror to cause the reflected ray to hit point R?

6 cm | |

12 cm | |

16 cm | |

18 cm |

Question 92 Explanation:

Triangle AXB is similar to triangle RCB. From the 2 similar triangles formed,
AX/XB = RC/CB => 8/x = 4/(24 – x)
x = 16 cm

Question 93 |

An observer O cannot see the image of target through the mirror. What is the minimum distance he should move to see the image os target T through the mirror?

1 m | |

2 m | |

3 m | |

4 m |

Question 93 Explanation:

Draw the image of T by using the extension of the mirror as the line of symmetry. Draw a line from image ‘T’ through the nearest edge of the mirror and extend the line to meet the horizontal line where the observer needs to move 2 m to the left to be able to see the image of T.

Question 94 |

Light passes through three different mediums as shown in the diagram below. Which of the following statements is true?

Medium 1 has the lowest refractive index. | |

Medium 2 has the highest refractive index. | |

The refractive index of medium 2 is lower than the refractive index of medium 3 | |

The refractive index of medium 1 is lower than the refractive index of medium 3 |

Question 94 Explanation:

For light ray travelling through parallel boundaries, the region with the highest refractive index will be the one with the smallest angle of refraction.

Question 95 |

A light ray passes through a rectangular glass block as shown in the diagram below. What is the refractive index of the glass?

sin 40 ÷ sin 60 | |

sin 60 ÷ sin 40 | |

sin 30 ÷ sin 50 | |

sin 50 ÷ sin 30 |

Question 95 Explanation:

The angle of incidence and the angle of refraction must be measured with respect to the normal.
sin i ÷ sin r = nt ÷ ni
ni = sin 50 ÷ sin 30

Question 96 |

A light ray passes through a rectangular glass block as shown in the diagram below. Which of the following expressions should be used to calculate the refractive index of the glass block?

sin 20 ÷ sin 50 | |

sin 50 ÷ sin 20 | |

sin 40 ÷ sin 20 | |

sin 20 ÷ sin 40 |

Question 96 Explanation:

The angle of incidence and the angle of refraction must be measured with respect to the normal.
Sin i ÷ sin r = nt ÷ ni
ni = sin 40 ÷ sin 20

Question 97 |

Refraction takes place when light travels from medium A into air as shown. What is the critical angle for the situation below?

48 deg | |

51 deg | |

55 deg | |

Critical angle cannot be reached |

Question 97 Explanation:

ni sin i = nt sin r
(ni) sin 30 = nt sin 40
ni = 1.29
ni sin i = nt sin r
(1.29) sin c = (1.0) sin 90
c = 51 deg.

Question 98 |

Light ray from air is directed towards point O (centre of the diameter) of the semi-curved glass block. Which diagram is not possible?

A | |

B | |

C | |

D |

Question 98 Explanation:

When light travels from glass to air (optically denser to less dense medium), light will not refract towards the normal. Choice C is a case of total internal reflection and choice D is a case of critical angle being reached.

Question 99 |

When is refraction of light not possible?

- The angle of incidence is 0 deg.
- The two mediums have the same refractive index.
- The refractive index is higher than 3.0.

1 and 2 only | |

1 and 3 only | |

2 and 3 only | |

1, 2 and 3 |

Question 99 Explanation:

1. The angle of incidence is 0 deg. (Correct)
When the angle of incidence is 0 deg, there is no bending of light ray. The angle of refraction is 0 deg and therefore the light ray travels straight through the mediums.
2. The mediums have the same refractive index. (Correct)
When the two mediums have the same refractive index, the light ray will not have a change speed and therefore will not bend at the boundary of the two mediums.
3. The refractive index is higher than 3.0 (Incorrect)
As long as the angle of incidence is not 0 deg and there is a difference between in the refractive index indices of the two mediums involved, refraction will take place.

Question 100 |

Which of the following is not the effect of refraction?

Chopsticks appear to bent in clear soup. | |

A fish appears to be larger than its actual size in water. | |

Light travelling through optical fibre. | |

A man appears to be smaller than his actual size from the point of view of a fish underwater. |

Question 100 Explanation:

When light travels through optical fibre, total internal reflection takes place and not refraction.

Question 101 |

What is/are the condition(s) needed for total internal reflection to take place?

- Light ray travels from an optically denser medium to an optically less dense medium.
- The angle of incidence is not 0 deg.
- The angle of incidence must be greater than the critical angle.

1 and 2 only | |

1 and 3 only | |

2 and 3 only | |

1, 2 and 3 |

Question 101 Explanation:

All the choices are conditions required to produce total internal reflection.

Question 102 |

Which of the following minimize the gap

*x*?- Use a thinner glass block
- Use a glass of smaller refractive index
- Use a stronger light ray

1 and 2 only | |

1 and 3 only | |

2 and 3 only | |

1, 2 and 3 |

Question 102 Explanation:

1. Use a thinner glass block. (Correct)
2. Use a glass of smaller refractive index. (Correct)
3. Use a stronger light ray. (Incorrect)
A stronger light ray can only make the emergent ray stronger. It cannot change the speed or the direction.

Question 103 |

An air bubble is trapped 6 cm from the edge of a glass block as shown. Given that the refractive index of glass is 1.5, how far does the air bubble seem to be away from the observer when viewed from the position shown?

4 cm | |

6 cm | |

9 cm | |

12 cm |

Question 103 Explanation:

Refractive index = Actual Depth / Apparent Depth
Actual Depth = 6 cm
1.5 = 6 / Apparent Depth
Apparent Depth = 6 cm / 1.5 = 4 cm

Question 104 |

A fish seems to be 6.0 cm below the water surface. Given that the refractive index of water is 1.4, what is the actual depth of the fish below the water?

4.3 cm | |

6.0 cm | |

8.4 cm | |

14.4 cm |

Question 104 Explanation:

Refractive index = Actual Depth / Apparent Depth
Apparent Depth = 6 cm
Actual Depth = Apparent Depth x 1.4
Actual Depth = 6 cm x 1.4 = 8.4 cm

Question 105 |

A | |

B | |

C | |

D |

Question 105 Explanation:

Incident light ray passing through the focal point must move parallel to the principal axis after passing through the converging lens. Therefore choice A, choice B and choice D cannot be the answer.

Question 106 |

A slide with an image 4 cm x 2 cm is placed at a distance of 10 cm behind a converging lens and a clear image is formed on a screen 1.1 m from the slide. The size of the image on the screen is

40 cm x 20 cm | |

20 cm x 40 cm | |

36 cm x 18 cm | |

18 cm x 36 cm |

Question 106 Explanation:

According to the geometry, the image is 10 times the size of the object. Since the object is 4 cm x 2 cm, the image must be 40 cm x 20 cm.

Question 107 |

Which of the following does not affect the focal length of a convex lens?

refractive index of the material making the lens | |

material of the lens | |

object distance | |

thickness of the lens |

Question 107 Explanation:

The object distance is the distance between the object and the optical centre. The object distance does not affect the focal length which is a property of the lens.

Question 108 |

a diverging lens | |

a converging lens | |

a glass block | |

a plane mirror |

Question 108 Explanation:

Converging lens and diverging lens will converge and diverge the rays respectively. A mirror will reflect the light. Only a rectangular glass block is able to produce such effect when light rays strike the glass block at 90 deg.

Question 109 |

A converging lens has a focal length of 20 cm. If an object is placed 50 cm from the optical centre of the lens, how far will the image be from the optical centre of the lens?

10 cm | |

20 cm | |

30 cm | |

33 cm |

Question 109 Explanation:

Use the lens equation,
1/u + 1/v = 1/f
Where u is the object distance, v is the image distance and f is the focal length.
1/50 + 1/v = 1/20
1/v = 1/20 – 1/50
v = 33.3 cm

Question 110 |

A converging lens has a focal length of 10 cm. If an object is placed 18 cm from the optical centre of the lens, how far will the image be from the optical centre of the lens?

20.0 cm | |

22.5 cm | |

22.5 cm | |

25.5 cm |

Question 110 Explanation:

Use the lens equation,
1/u + 1/v = 1/f
Where u is the object distance, v is the image distance and f is the focal length.
1/18 + 1/v = 1/10
1/v = 1/10 – 1/18
v = 22.5 cm

Question 111 |

A converging lens has a focal length of 10 cm. If an object is placed 5 cm from the optical centre of the lens, how far from the optical centre of the lens will the virtual image appear?

5.0 cm | |

10.0 cm | |

15.0 cm | |

18.5 cm |

Question 111 Explanation:

Use the lens equation,
1/u + 1/v = 1/f
Where u is the object distance, v is the image distance and f is the focal length.
1/5 + 1/v = 1/10
1/v = 1/10 – 1/5
v = – 10 cm
A negative answer indicates that the image is on the same side as the object. It also implies that the image is a virtual image.

Question 112 |

A converging lens has a focal length of 10 cm. If a real image is formed 20 cm away from the optical centre, how far away is the object placed from the optical centre of the lens?

10 cm | |

15 cm | |

20 cm | |

25 cm |

Question 112 Explanation:

Use the lens equation,
1/u + 1/v = 1/f
Where u is the object distance, v is the image distance and f is the focal length.
1/u = 1/10 – 1/20
u = 20 cm
The answer can also be obtained by substituting u = v = 2f and solving for u.

Question 113 |

A converging lens has a focal length of 10 cm. If a virtual image is formed 20 cm away from the optical centre, how far away is the object placed from the optical centre of the lens?

5.0 cm | |

6.7 cm | |

10 cm | |

20 cm |

Question 113 Explanation:

Use the lens equation,
1/u + 1/v = 1/f
Where u is the object distance, v is the image distance and f is the focal length.
1/u = 1/10 + 1/20
u = 6.7 cm

Question 114 |

In an experiment to determine the focal length of a converging lens, the object distance and the image distance recorded were 20 cm and 30 cm respectively. What is the focal length of the converging lens?

10 cm | |

12 cm | |

15 cm | |

18 cm |

Question 114 Explanation:

Use the lens equation,
1/u + 1/v = 1/f
Where u is the object distance, v is the image distance and f is the focal length.
1/f = 1/20 + 1/30
f = 12 cm

Question 115 |

A student used a converging lens, light source and a transparency to produce an image on the screen. He found out that the picture on the transparency was 5 cm and the image formed on the screen was 25 cm. If the screen is 3 m away from the lens, how far away will the transparency be from the lens?

10 cm | |

50 cm | |

60 cm | |

1500 cm |

Question 115 Explanation:

Based on the principle of similar triangles,
x/300 = 5/25
x = 60 cm

Question 116 |

A converging lens is used to magnify a real image to three times its original size. Given that the object is placed 20 cm from the converging lens, what is the image distance?

10 cm | |

6.7 cm | |

60 cm | |

80 cm |

Question 116 Explanation:

Based on the principle of similar triangles,
x = 60 cm

Question 117 |

A converging lens is used to magnify a real image to four times its original size. Given that the focal length of the converging lens is 20 cm, what is the object distance?

5 cm | |

25 cm | |

40 cm | |

100 cm |

Question 117 Explanation:

Use the lens equation,
1/u + 1/v = 1/f
Where u is the object distance, v is the image distance and f is the focal length.
From the diagram,
1/u = 4/v
v = 4u
1/u + 1/4u = 1/20
5/4u = 1/20
u = 25 cm

Question 118 |

A converging lens is used to magnify a virtual image to four times its original size. Given that the focal length of the converging lens is 20 cm, what is the object distance?

5 cm | |

10 cm | |

15 cm | |

25 cm |

Question 118 Explanation:

Use the lens equation,
1/u + 1/v = 1/f
Where u is the object distance, v is the image distance and f is the focal length.
From the diagram,
1/u = 4/v
v = 4u
1/u + 1/-4u = 1/20
3/4u = 1/20
u = 15 cm

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