## Acceleration due to gravity – problems and solutions

1. What is the acceleration due to gravity at the Moon’s surface? Moon’s mass = 7.35 x 1022 kg, the radius of the Moon is 1.7 x 106 m, universal constant is 6.67 x 10-11 N m2 / kg2

Solution

The formula of Newton’s second law of motion :

The formula of Newton’s law of universal gravitation :

G = universal constant, M = Earth’s mass, m = object’s mass, r = the distance between Earth’s center and object. If the object is at Earth’s surface, r = Earth’s radius

Substitute F in equation 1 with F in equation 2 :

What is the acceleration due to gravity at the Moon’s surface?

[irp]

2. Acceleration due to gravity at Earth’s surface is g. At R (R is Earth’s surface), the acceleration due to gravity is…

Solution

R = Earth’s radius. At R above Earth’s surface = at 2R above Earth’s center.

At R above Earth’s surface, the acceleration due to gravity = ¼ g. If g = 9.8 m/s2, at R above Earth’s surface, acceleration due to gravity = 2.45 m/s2.

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## Gravitational force Weight – Problems and Solutions

5 Gravitational force Weight – Problems and Solutions

1. What is the force of gravity acting on an object at the Earth’s surface? Earth’s mass = 5.98 x 1024 kg, object’s mass = 1000 kg, the radius of the Earth is 6.38 x 106 m.

Known :

Earth’s mass (mE) = 5.98 x 1024 kg

Object’s mass (mo) = 1000 kg

The radius of the Earth (rE) = 6.38 x 106 m

Universal constant (G) = 6.67 x 10-11 N m2 / kg2

Acceleration due to gravity (g) = 9.8 m/s2

Wanted : the force of gravity

Solution :

w = weight, F = force, G = universal constant, mE = Earth’s mass, mo = object’s mass, r = the distance between the Earth’s center and object.

The object is at the surface of the Earth, so r = the radius of the Earth

Object’s weight (Newton’s second law of motion) :

w = m g

w = (1000 kg)(9.8 m/s2)

w = 9800 N

[irp]

2. What is the force of gravity acting on an object at 10,000 meters above the Earth’s surface ? Earth’s mass = 5.98 x 1024 kg, object’s mass = 1000 kg, the radius of the Earth is 6.38 x 106 m.

Solution :

3. The weight of a spacecraft is w. If D = Earth’s diameter, determine the weight of spacecraft when the spacecraft is at 2D above the Earth’s surface.

Known :

D = Earth’s diameter,

R = the radius of the Earth

1 D = 2 R, 2 D = 4 R

Wanted: spacecraft’s weight at 2D above the Earth’s surface?

Solution :

[irp]

4. The ratio of the mass of the planet A and planet B is 2 : 3, while the ratio of the radius of the planet A and planet B is 1 : 2. If the weight of an object on planet A is w, what is the weight of the object on the planet B.

Known :

Mass of planet A (mA) = 2

Mass of planet B (mB) = 3

Radius of the planet A (rA) = 1

Radius of planet B (rB) = 2

Mass of object = m

Object’s weight on planet A = w

Wanted: Object’s weight on planet B

Solution :

The equation of the force of weight from Newton’s law of gravity :

w = weight, G = gravitational constant, M = mass of planet, m = mass of object, r = the distance between object and planet. If the object on the planet surface the r = radius of the planet.

Object’s weight on planet A :

Object’s weight on planet B :

Object’s mass is same so that substitute m with w/2G.

[irp]

5. A rocket with the weight of w launched vertically from the surface of the Earth. D is the diameter of the Earth. When rocket at the height of 0.5 D above the surface of the Earth, then what is the weight of the rocket.

Known :

Rocket’s weight = w

The radius of Earth = distance from the center of Earth = R

Diameter of Earth = D = 2R

Wanted: Weight of rocket when the rocket at the height of 0.5 D above the Earth surface.

Solution :

0.5 D = 0.5 (2R) = R

The distance of rocket from the center of Earth = radius of Earth + radius of a rocket from the surface of Earth = R + R = 2R

Weight is the force of gravity that acts on an object. The force of gravity (F) is inversely proportional to the square of the distance from the center of the earth (R) so that the weight is inversely proportional to the square of the distance.

## Optical instrument telescopes – problems and solution

1. The focal length of the objective lens is 400 cm and the focal length of the ocular lens is 20 cm. If the eye is relaxed, determine :

a) The overall magnification of telescope (M)

b) the image distance from the objective lens (dob’),

c) the image distance from the ocular lens (doc)

d) the distance between the lenses

Known :

The focal length of objective lens (fob) = 400 cm

The focal length of ocular lens (foc) = 20 cm

The eye is relaxed so that the real image generated by the objective lens is at the focal point of the objective lens and the first focal point of the ocular lens.

Solution :

a) The overall magnification (M)

M = fob / foc

M = 400 cm / 20 cm = 20X

The image is virtual and inverted.

b) The image distance from objective lens (dob’)

If the eye is relaxed, the image distance from objective lens (dob’) = the focal length of objective lens (fob) = 400 cm = 4 m.

c) The distance of real image from the ocular lens (doc)

If the eye is relaxed, the distance of real image and ocular lens (soc) = the focal length of ocular lens (foc) = 20 cm = 0.2 m.

d) The distance between the lenses

The distance between the lenses = the telescope length (l) = the focal length of objective lens (fob) + the focal length of ocular lens (foc) = 400 cm + 20 cm = 420 cm.

[irp]

2. The distance between objective lens and ocular lens is 20 cm. The overall magnification of telescope when the eye is relaxed is 40X. Calculate the focal length of ocular lens (foc) and the focal length of objective lens (fob)

Known :

the telescopes length (l) = 20 cm

the overall magnification of telescope (M) = 40X

Solution :

If the eye is relaxed, the focal length of ocular lens (foc) + the focal length of objective lens (fob) = the telescope length (l)

foc + fob = l

fob = l – foc

fob = 20 – foc —–> equation 1

The overall magnification of telescope (M) :

M = fob / foc

fob = M foc

fob = 40 foc —–> equation 2

The focal length of ocular lens :

Subtitute fob in equation 1 with fob in equation 2 :

fob = 20 – foc

40 foc = 20 – foc

40 foc + foc = 20

41 foc = 20

foc = 20 cm / 41

foc = 0.5 cm

The focal length of ocular lens = 0.5 cm.

The focal length of the objective lens :

fob = 20 – foc

fob = 20 – 0.5

fob = 19.5 cm

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## Optical instrument microscope – problems and solutions

1. The focal length of the objective lens is 2 cm and the focal length of the ocular lens is 5 cm. The distance between the objective lens and the ocular lens is 30 cm. Determine the overall magnification and the object distance from the objective lens when the eye is relaxed.

Known :

The focal length of objective lens (fob) = 2 cm

The focal length of ocular lens (foc) = 5 cm

Distance between the lenses (l) = 30 cm

Near point (N) = 25 cm

Solution :

(a) the overall magnification

The formula of the overall magnification :

M = mob Moc

M = the overall magnification, mob = the magnification of the objective lens, Moc = the ocular angular magnification

The magnification of the objective lens when the eye is relaxed (mob) :

The image distance from the objective lens (dob’) :

dob’ = l – foc = 30 cm – 5 cm = 25 cm

The object distance from the objective lens (dob) :

The objective lens is a converging lens so the focal length has the plus sign.

The image distance has plus sign because the image is real or the rays pass through the image.

The magnification of the objective lens :

The ocular angular magnification when the eye is relaxed (Mok) :

Moc = N / foc = 25 cm / 5 cm = 5X

The overall magnification :

M = mob Moc = (12.5)(5) = 62.5X

(b) The object distance from objective lens (dob)

The object distance from the objective lens is 2 cm.

[irp]

2. A microscope consists of a 5X objective and a 20X ocular. The distance between the lenses is 15 cm. (a) Determine the overall magnification if the eye is relaxed (b) determine the focal length of the ocular lens (c) the focal length of the objective lens

Known :

The magnification of the objective lens (mob) = 5X

The magnification of ocular lens (Moc) = 20X

Near point (N) = 25 cm

The distance between the lenses (l) = 15 cm

Solution :

(a) The overall magnification (M)

M = mob Moc

M = (5)(20) = 100X

(b) The focal length of the ocular lens (foc)

The formula of angular magnification of ocular lens (Moc) when the eye is relaxed :

Moc = N / foc

The focal length of the ocular lens :

foc = N / Moc = 25 cm / 20 = 1.25 cm

(c) the focal length of the objective lens (fob)

The formula of the magnification of the objective lens when the eye is relaxed :

The distance of the real image from the objective lens (dob’) :

dob’ = l – foc = 15 cm – 1.25 cm = 13.75 cm

The object distance from the objective lens (dob) :

The focal length of the objective lens (fob) :

The focal length of the objective lens = 2.29 cm

3. The focal length of the objective lens is 0.9 cm and the focal length of the ocular lens is 2.5 cm. The microscope is used by a normal eye without accommodation and the overall magnification is 90 times. Determine the distance between the object and the objective lens. N = 25 cm.

Known :

The focal length of the objective lens (fob) = 0.9 cm

The focal length of the ocular lens (fok) = 2,5 cm

The overall magnification (M) = 90 times

The near point of a normal eye (N) = 25 cm

The eye’s accommodation is minimum.

Wanted: The distance between the object and the objective lens (sob)

Solution :

The equation of the total angular magnification when the accommodation is minimum :

Calculate the object distance about the objective lens (sob) using the equation of relation between the focal length of the objective lens (fob), the distance between the object and the objective lens (sob) and the distance between the image and the objective lens (sob‘) :

4. The focal length of the objective lens is 1.8 cm and the focal length of the ocular lens is 6 cm. The microscope is used by a normal eye without accommodation, the distance between the objective lens and the ocular lens is 24 cm. Determine the object distance from the objective lens.

Known :

The focal length of the objective lens (fob) = 1.8 cm

The focal length of the ocular lens (fok) = 6 cm

The distance between the objective lens and the ocular lens = the length of the microscope (l) = 24 cm

The accommodation is a minimum.

Wanted: Distance between the object and the objective lens (sob)

Solution :

When the accommodation is minimum, the distance between the final image and the ocular lens is infinity, as shown in the figure below.

The distance between the objective lens and the ocular lens (l) = the focal length of the ocular lens (fok) + the distance between the image and the objective lens (sob’).

sob’ = l – fok = 24 cm – 6 cm = 18 cm

Calculate the distance between the object and the objective lens (sob) using the equation of relation between the focal length of the objective lens (fob), the distance between the object and the objective lens (sob) and the distance between the image and the objective lens (sob‘) :

The distance between the object and the objective lens is 2 cm.

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## Optical instrument magnifying glass – problems and solutions

1. A 2 mm high object is placed 10 cm from a magnifying glass. Near point N = 25 cm. Determine the angular magnification and image height.

Known :

Object height (ho) = 2 mm

Near point (N) = 25 cm

Object distance (do) = 10 cm

Wanted : Angular magnification (M) and image height (hi)

Solution :

M = N / s

M = 25 cm / 10 cm

M = 2.5

The image height = 2.5 x 2 mm = 5 mm.

2. A 25 cm focal length lens is used as magnifying glass. Determine (a) angular magnification when the eye is focused at its near point N = 25 cm (b) angular magnification when the eye is relaxed.

Known :

Near point (N) = 25 cm

The focal length of a magnifying glass (f) = 25 cm

Solution :

(a) angular magnification when the eye is focused at its near point N = 25 cm

M = (N/f) + 1

M = (25 cm / 25 cm) + 1

M = 1 + 1

M = 2 X

If the object height is 1 cm, the image height is 2 x 1 cm = 2 cm.

(b) angular magnification when the eye is relaxed

M = N / f

M = (25 cm / 25 cm)

M = 1 X

If the object height is 1 cm, the image height is 1 x 1 cm = 1 cm.

3. A 1 cm high object is placed in front of a 10 cm focal length lens. Determine (a) the image height when the eye is focused at its near point N = 25 cm (b) The image height when the eye is relaxed.

Known :

The object height (ho) = 1 cm

The focal length (f) = 10 cm

Near point (N) = 25 cm

Solution :

(a) The image height when the eye is focused at its near point N = 25 cm

M = (N/f) + 1

M = (25 cm / 10 cm) + 1

M = 2.5 + 1

M = 3.5 X

If the object height is 1 cm, the image height is 3.5 x 1 cm = 3.5 cm.

(b) The image height when the eye is relaxed.

M = N/f

M = 25 cm / 10 cm

M = 2.5 X

If the object height is 1 cm, the image height is 2.5 x 1 cm = 2.5 cm.

4. The angular magnification when the eye is relaxed = 5X. If near point = 25 cm, what is the focal length of the magnifying glass ?

Known :

Object height (ho) = 2 mm

Angular magnification (M) = 5X

Near point (N) = 25 cm

Wanted: The focal length

Solution :

The formula of the angular magnification when the eye is relaxed :

M = N/f

5 = 25 cm / f

f = 25 cm / 5

f = 5 cm

The focal length of the magnifying glass = 5 cm.

5. An object is seen by someone with a magnifying glass with the focal length is 15 cm. If the near point of the person’s eyes = 30 cm, then determine the overall magnification of the magnifying glass.

Known :

The near point of the normal eye (N) = 30 cm

The focal length of the magnifying glass (f) = 15 cm (plus sign because the glass is convergent)

Wanted : the maximum magnification

Solution :

The maximum magnification occurs when the accommodation of eye is maximum. The angular magnification of the magnifying glass occurs when the accommodation of eye is maximum :

M = (N/f) + 1

M = (30 cm / 15 cm) + 1

M = 2 + 1

M = 3 times

6. A magnifying glass with the optical power 20 diopters used by a person with the normal eyes 25 cm. If the accommodation is minimum, determine the minimum magnification.

Known :

Near point of the normal eye (N) = 25 cm

Power of the magnifying glass (P) = 20 diopters

Wanted: The minimum magnification

Solution :

The focal length of the magnifying glass :

P = 1/f

20 = 1/f

f = 1/20

f = 0.05 meters

f = 5 cm

The angular magnification when the accommodation is minimum :

M = N / f

M = (25 cm / 5 cm)

M = 5 times

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## Optical instrument eyeglasses – problems and solutions

1. Power of an eyeglass lens is -2 Diopters. The distance between the eye and eyeglass is 2 cm.

(a) The eyeglass has a converging lens or diverging lens?

(b) What is the focal length of the lens?

(c) Nearsighted eye or farsighted eye? If farsighted, what then will be the far point?

Known :

Lens power (P) = -2 Diopters

Solution :

(a) Converging lens or diverging lens

The minus sign of the lens’ power indicates the contact lens is diverging lens.

(b) The focal length

P = 1/f

-2 = 1/f

f = 1/-2 = -0.5 m = -50 cm

The focal length of diverging lens is 50 cm.

(c) Nearsighted eye or farsighted eye ?

-1/di = 1/f – 1/do

-1/di = -1/50 – 1/~ = -1/50 – 0

-1/di = -1/50

di = 50 cm

The image distance is 50 cm + 2 cm = 52 cm. 52 cm is far point of the farsighted eye. For a normal eye, the far point is infinity.

[irp]

2. Power of a eyeglass lens is 4 Diopters. The distance between eye and eyeglass is 2 cm.

(a) The eyeglass has converging lens or diverging lens ?

(b) What is the focal length of the lens ?

eye ? If nearsighted, what then will be the near point ?

Known :

lens power (P) = 4 Diopters

Solution :

(a) Converging lens or diverging lens

The plus sign of the lens’ power indicates the contact lens is converging lens.

(b) The focal length

P = 1/f

4 = 1/f

f = 1/4 = 0.25 m = 25 cm

The focal length of converging lens is 25 cm.

(c) Nearsighted eye or farsighted eye ?

-1/di = 1/f – 1/do

-1/di = 1/25 – 1/23 = 23/575 – 25/575 = -2/575

di= 575/-2 = -287.5 cm = -2.875 meters

di = 287.5 cm = 2.875 meters

The image distance is 287.5 cm + 2 cm = 289.5 cm. 289.5 cm is near point of the nearsighted eye. For a normal eye, the near point is 25 cm.

3. An eye has a near point 16 cm and a far point of 80 cm. If he uses glasses, he can see far objects clearly. Using the eyeglass, distance of the closest object that can be seen clearly is ….

A. 13 1/3 cm

B. 20 cm

C. 36 cm

D. 48 1/3 cm

Known :

The far point of the person is 80 cm, therefore, concluded that the person is suffering nearsightedness The nearsightedness caused by the eye lens is more curved than it should be like in the normal eye so the focal length of the eye lens is reduced. This causes the beam of light from the infinite (far point) not focus on the retina but focused in the front of the retina.

Wanted: Distance of the closest thing that can be seen clearly using the glasses

Solution :

The person’s far point is 80 cm. The lens of the eyeglass should produce an image at a distance of 80 cm in front of it. The image is in front of the eyes and the lens of the glasses so that the image is virtual and upright. So the distance of the image (d’) = -80 cm. If the person is wearing eyeglass, he can see very far objects clearly. So the distance of the object (d) = the far point of the normal eye = infinity = ~.

The focal length of the eye lens :

1/f = 1/d + 1/d’

1/f = 1/~ + (- 1/80)

1/f = 0 – 1/80

1/f = – 1/80

f = – 80 / 1

f = – 80 cm

The focal length signed negative means the lens of eyeglass used is a concave lens or diverging lens.

If the person is using the same eyeglass lens then how close is the closest thing that can be seen clearly? Focal length of lens (f) = -80 cm. The lens should produces an image at a distance of 16 cm in front of the eye and the lens, so that the image is virtual and signed negative. So the distance of the image (d’) = -16 cm.

1 / d = 1 / f – 1 / d’ = -1/80 – (-1/16) = -1/80 + 1/16 = -1/80 + 5/80 = 4/80

d = 80/4 = 20 cm.

The closest object can be seen clearly is 20 cm.

4. Based on the figure, can be concluded that…

Solution :

 Farsighted eye Farsighted eye + converging lens

Converging lens = convex lens = positive lens

5. The near point of a hypermetropy sufferer is 2 m. In order to see a normal-eyed person then the person needs to use eyeglass…

A. -1.5 Diopters

B. -2.5 Diopters

C. +3.5 Diopters

D. +4.5 Diopters

Known :

The near point of hypermetropy sufferers (farsightedness) = 2 meters

The near point of the normal eye = 25 cm = 0.25 meters

Wanted: The power of lens glasses

Solution :

The lens should produce an image at a distance of 2 meters in front of the eye so that the image is virtual and signed negative. So the distance of image (d’) = – 2 meters.

In order to see like a normal-eye person, the distance of the object (d) = the near point of the normal eye = 25 cm = 0.25 meters.

The focal length of the lens of eyeglass :

1/f = 1/d + 1/d’

1/f = 1/0.25 + (-1/2) = 1/0.25 – 1/2 = 8/2 – 1/2 = 7/2

f = 2/7

The focal length signed positive means that the lens of glass used is a positive lens or a convex lens or a converging lens.

The power of the lens:

P = 1 / f = 1: 2/7 = 1 x 7/2 = 7/2 = +3.5 Diopters

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## Optical instrument contact lenses – problems and solutions

1. Power of a contact lens is -5 Diopters.

(a) The contact lens is a converging lens or diverging lens?

(b) What is the focal length of the contact lens?

(c) Nearsighted eye or farsighted eye? If the nearsighted eye, what then will be the near point?

Known :

Lens’ power (P) = -5 D

Solution :

(a) Converging lens or diverging lens ?

The minus sign of the lens’ power indicates the contact lens is the diverging lens.

(b) The focal length?

P = 1/f

-5 = 1/f

f = 1/-5 = -0.2 m = -20 cm

The focal length of diverging lens is 20 cm.

(c) Nearsighted eye or farsighted eye ?

-1/di = 1/f – 1/do

-1/di = -1/20 – 1/~ = -1/20 – 0

-1/di = -1/20

di = 20 cm

The image distance is 20 cm. 20 cm is far point of the farsighted eye. For a normal eye, the far point is infinity.

[irp]

2. Power of a contact lens is 1.5 Diopters.

(a) The contact lens is converging lens or diverging lens ?

(b) What is the focal length of the contact lens ?

(c) Nearsighted eye or farsighted eye ? If farsighted eye, what then will be the far point ?

Known :

Lens power (P) = 1.5 Diopters

Solution :

(a) converging lens or diverging lens

The plus sign of the lens’ power indicates the contact lens is converging lens.

(b) The focal length

P = 1/f

1.5 = 1/f

f = 1/ 1.5 = 0.67 m = 67 cm

The focal length of converging lens is 67 cm.

(c) Nearsighted eye or farsighted eye

-1/di = 1/f – 1/do

-1/di = 1/67 – 1/25 = 25/1675 – 67/1675 = -42/1675

di = -1675/42 = -40 cm = -0.40 m

di = 40 cm

The image distance is 40 cm. 20 cm is near point of the farsighted eye. For a normal eye, the near point is 25 cm.

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## Optical instrument human eye – problems and solutions

1. A normal eye has a far point of infinity. The focal length of the eye lens is 2.5 cm. Determine (a) the image distance (b) lens power.

Known :

The focal length of the eye lens (f) = the distance between the cornea and retina = +2.5 cm (The plus sign indicates the converging lens).

The object distance (do) = infinity

Wanted : the image distance (di)

Solution :

(a) the image distance

1/f = 1/do + 1/di

f = the focal length, do = the object distance, di = the image distance

1/di = 1/f – 1/do = 1 / 2.5 – 1 / ~ = 1 / 2.5 – 0

1/di = 1 / 2.5

di = 2.5 cm = 0.025 meter

If the object distance is infinity, the image distance (di) = the focal length (f)

(b) lens power

P = 1/f = 1/ 0.025 m = 40 Diopters

[irp]

2. A normal eye has a near point of 25 cm. The image distance is the same as the distance between cornea and retina = 2.5 cm. Determine (a) the focal length (b) lens power

Known :

The object distance (do) = 25 cm

The image distance (di) = 2.5 cm

Solution :

(a) the focal length (f)

1/f = 1/do + 1/di

1/f = 1/ 25 + 1/ 2,5 = 1/25 + 10/25 = 11/25

f = 2.27 cm = 0.0227 m

(b) lens power (P)

P = 1/f = 1/ 0.0227 = 44 Diopters

3. One of the optical instrument is human eyes. Determine the properties of image produced by human eyes…

A. Real, upright

B. Real, inverted

C. Virtual, upright

D. Virtual, inverted

Solution

The near point of the normal eye is 25 cm. The near point is the point nearest the eye at which an object is accurately focused on the retina at the full accommodation. The diameter of the human eye is smaller than 25 cm so the focal length of the lens of the eye is also smaller than 25 cm. Thus, the distance of the object (d) must be greater than the focal length of the lens of the eye (f).

The eye lens is a convex lens. Therefore the image formed by the eye lens has the same properties as the image formed by a convex lens. The properties of an image formed by the eye lens when the object distance (s) is greater than the focal length (f) :

real, inverted, smaller

The image should be focused at the retina because the retina converts the light waves into electrical signals transmitted to the brain. Although the image is inverted the human brain turns it upright so that the objects seen by the eye are upright.

When you see a tree from a distance, the tree looks smaller than its actual size. This is in accordance with the properties of the image that is reduced.

4. The eye can see an object when the image is focused on the retina where the properties of the image are…

A. Virtual, upright, greater

B. Virtual, inverted, smaller

C. Real, inverted, smaller

D. Real, upright, smaller

Solution
– The lens of the eye is a convex lens, therefore, the properties of an image formed by the eye lens are similar to the properties of an image formed by the convex lens.

– The near point of the normal eye or the closest distance that can still be seen clearly by the eyes is 25 cm. The eyeballs are so small so the radius of curvature and the focal length of the eye lens is smaller than 25 cm. So can be concluded that the object distance is greater than the focal length.

– If the object distance is larger than the focal length then the properties of an image formed by the human eye at the retina are real, inverted, and smaller. Although the image is inverted, the brain turns it upright as we experience it.

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## Converging lens – problems and solutions

1. A 5-cm high object is placed 5 cm from a 15-cm focal length converging lens. Determine the image distance, the magnification of the image, the image height and the properties of the image.

Known :

The focal length (f) = 15 cm

The plus sign indicates that the focal point is real or the rays pass through the point.

The object height (ho) = 5 cm

The object distance (do) = 5 cm

Wanted : the image distance, the magnification of image, the image height and the properties of image

Solution :

Formation of image by converging lens :

The image distance (di) :

1/di = 1/f – 1/do = 1/15 – 1/5 = 1/15 – 3/15 = -2/15

di = -15/2 = -7.5 cm

The minus sign indicates that the image is virtual or the rays do not pass through the image.

The magnification of image (m) :

m = – di / do = -(-7.5)/5 = 7.5/5 = 1.5

The image height (hi) :

m = hi / ho

hi = m ho = (1.5)5 = 10/3 = 7.5 cm

The properties of the image

virtual

upright

The image greater than the object

The image distance is greater than the object distance

[irp]

2. A 10-cm high object is placed 30 cm from a 15-cm focal length converging lens. Determine the image distance, the magnification of the image, the image height and the properties of the image.

Known :

The focal length (f) = 15 cm

The plus sign indicates that the focal point is real or the rays pass through the point.

The object height (h) = 10 cm

The object distance (s) = 30 cm

Wanted: The image distance, the magnification of the image, the image height and the properties of the image

Solution :

Formation of the image by the converging lens :

The image distance (di) :

1/di = 1/f – 1/do = 1/15 – 1/30 = 2/30 – 1/30 = 1/30

di = 30/1 = 30 cm

The plus sign indicates that the image is real or the rays pass through the image.

The magnification of image (m) :

m = – di / do = -(30)/30 = -30/30 = -1

The image height (hi) :

m = hi / ho

hi = m ho = (-1)10 = -10 cm

The properties of the image

real or the rays pass through the image

inverted

The image height is the same as the object height.

The image distance is the same as the object distance.

[irp]

3. A 10-cm high object is placed 30 cm from a 20-cm focal length converging lens. Determine the image distance, the magnification of the image, the image height and the properties of the image.

Known :

The focal length f) = 20 cm

The plus sign indicates that the focal point is real or the rays pass through the point.

The object height (h) = 10 cm

The object distance (do) = 30 cm

Wanted : The image distance, the magnification of image, the image height and the properties of image

Solution :

The image distance (di) :

1/di = 1/f – 1/do = 1/20 – 1/30 = 3/60 – 2/60 = 1/60

di = 60/1 = 60 cm

The plus sign indicates that the image is real or the rays pass through the image.

The magnification of image (m) :

m = – di / do = -(60)/30 = -2

The image is 2 times greater than the object.

The image height (hi) :

m = hi / ho

hi = m ho = (-2)10 = -20 cm

The properties of the image

real

inverted

the image height is greater than the object height

the image distance is greater than the object distance

[irp]

4. Based on the figure below, determine the focal length of the converging lens!

Known :

Object distance (do) = 20 cm

Image distance (di) = 30 cm

Wanted : The focal length (f)

Solution :

1/do + 1/di = 1/f

do = object distance (plus sign because rays pass through object)

di = image distance (plus sign because rays pass through image)

f = focal length (plus sign because rays pass through the focal point or the focal point is real)

The focal length :

1/f = 1/20 + 1/30 = 3/60 + 2/60 = 5/60

f = 60/5 = 12 cm

5. Object’s distance is 30 cm and the focal length is 20 cm. Determine the magnification of the image.

Known :

The focal length of the converging lens (f) = 20 cm

The plus sign indicates that the focal point is real or rays pass through the image.

Object distance (do) = 30 cm

Wanted : The magnification of image (M)

Solution :

Image distance :

1/di = 1/f – 1/do = 1/20 – 1/30 = 3/60 – 2/60 = 1/60

di = 60/1 = 60 cm

The magnification of image :

M = di / do = 60 cm / 30 cm = 60/30 = 2 times

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## Diverging lens – problems and solutions

1. A 5-cm high object is placed 15 cm from a 30-cm focal length diverging lens. Determine the image distance, the magnification of the image, the image height, and properties of the image.

Known :

The focal length (f) = -30 cm

The minus sign indicates that the focal point is virtual or the rays do not pass through the point.

The object height (ho) = 5 cm

The object distance (do) = 15 cm

Wanted: The image distance (di), the magnification of image (m), the image height (hi) and the properties of the image.

Solution :

Formation of image by diverging lens :

The image distance (di) :

1/di = 1/f – 1/do = -1/30 – 1/15 = -1/30 – 2/30 = -3/30

di = -30/3 = -10 cm

The minus sign indicates that the image is virtual or the rays do not pass through the image.

The magnification of image (m) :

m = – di / do = -(-10)/15 = 10/15 = 2/3

The image 2/3 smaller than the object.

The image height (hi) :

m = hi / ho

hi = m ho = (2/3)5 = 10/3 = 3.3 cm

The properties of the image :

The properties of the image formed by a diverging mirror :

– virtual

– upright

– the image smaller than the object

– the image distance smaller than the object distance

[irp]

2. A 10-cm high object is placed 60 cm from a 30-cm focal length diverging lens. Determine the image distance, the magnification of the image, the image height, the properties of the image.

Known :

The focal length (f) = -30 cm

The minus sign indicates that the focal point is virtual or the rays do not pass through the point.

The object height (h) = 10 cm

The object distance (do) = 60 cm

Wanted : The image distance (di), the magnification of image (m), the image height (hi) and the properties of image

Solution :

Formation of image by diverging lens :

The image distance (di) :

1/di = 1/f – 1/do = -1/30 – 1/60 = -2/60 – 1/60 = -3/60

di = -60/3 = -20 cm

The minus sign indicates that the image is virtual or the rays do not pass through the image.

The magnification of image (m) :

m = -di/do = -(-20)/60 = 20/60 = 1/3

The image 1/3 smaller than the object.

Th image height (hi) :

m = hi / ho

hi = m ho = (1/3)10 = 10/3 = 3.3 cm

Properties of image

– virtual because the rays do not pass through the image

– upright

– the image smaller than the object

– the image distance smaller than the object distance

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