Physics

1. The Earth’s distance from the Sun is 149.6 x 10⁶ km and period of Earth’s revolution is 1 year. Calculate T² / r³

Known :

T = 1 year, r = 149.6 x 10⁶ km

Wanted : T² / r³ = … ?

Solution :

k = T² / r³ = 1² / (149.6 x 10⁶)³ = 1 / (3348071.9 x 10¹⁸) = 2.98 x 10^-25 year²/km³

2. Universal constant (G) = 6.67 x 10^-11 N.m²/kg² and Sun’s 1.99 x 10³⁰ kg.

3. The mean distance of Earth from the Sun is 149.6 x 10⁶ km and the mean distance of Mercury from the Sun is 57.9 x 10⁶ km. The period of Earth’s revolutions is 1 year, what is the period of Mercury’s revolution?

Known :

r of Earth = 149.6 x 10⁶ km

r of mercury = 57.9 x 10⁶ km

T of Earth = 1 year

Wanted: T of mercury?

Solution :

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1. Newton’s law of universal gravitation problems and solutions
2. Gravitational force, weight problems, and solutions
3. Acceleration due to gravity problems and solutions
4. Geosynchronous satellite problems and solutions
5. Kepler’s law problems and solutions problems and solutions

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1. What is the height of a geosynchronous satellite above Earth’s surface?

G = universal constant = 6.67 x 10^-11 N m²/kg²

M = Earth’s mass = 5.97 x 10²⁴ kg

T = 24 hours = 24 (3600 seconds) = 86,400 seconds = 8.64 x 10⁴ seconds

Solution

Satellite’s height above Earth’s surface = 4.22 x 10⁷ m = 4.22 x 10⁴ km = 42,200 km

Satellite’s height above Earth’s surface = 6.38 x 10⁶ m = 6.38 x 10³ km = 6380 km

Satellite’s height above Earth’s surface = satellite’s distance from Earth’s center – Earth’s radius

Satellite’s height above Earth’s surface = 42,200 km – 6380 km

Satellite’s height above Earth’s surface = 35,820 km

2. What is the speed of a geosynchronous satellite ?

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

Earth’s mass (m_E) = 5.97 x 10²⁴ kg

The distance between satellite and Earth’s center (r) = 4.22 x 10⁷ m

Solution

or use this equation

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1. Newton’s law of universal gravitation problems and solutions
2. Gravitational force, weight problems, and solutions
3. Acceleration due to gravity problems and solutions
4. Geosynchronous satellite problems and solutions
5. Kepler’s law problems and solutions problems and solutions

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1. The distance between a 40-kg person and a 30-kg person is 2 m. What is the magnitude of the gravitational force each exerts on the other. Universal constant = 6.67 x 10^-11 N m² / kg²

Known :

m₁ = 40 kg, m₂ = 30 kg, r = 2 m, G = 6.67 x 10^-11 N m² / kg²

Wanted : the magnitude of gravitational force (F) ?

Solution :

2. The distance between Earth and Moon (r) is 3.84 x 10⁸ m. What is the magnitude of gravitational force each exerts on the other?

Known :

Earth’s mass (m_E) = 5.97 x 10²⁴ kg

Moon’s mass (m_M) = 7.35 x 10²² kg

The distance between Earth’s center and the Moon’s center (r) = 3.84 x 10⁸ m

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

Wanted: The magnitude of gravitational force (F)

Solution :

3. What is the distance from the Earth, the magnitude of the gravitational force of the earth and the moon is zero?

Known :

Earth’s mass = 5.97 x 10²⁴ kg

Moon’s mass = 7.35 x 10²² kg

The distance between Earth’s center and Moon’s center (r) = 3.84 x 10⁸ m.

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

Wanted: distance from Earth

Solution :

Net gravitational force is zero.

Use quadratic formula :

A = – 6.35

B = – 7.68 x 10⁸

C = 87.76 x 10^-14

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1. Newton’s law of universal gravitation problems and solutions
2. Gravitational force, weight problems, and solutions
3. Acceleration due to gravity problems and solutions
4. Geosynchronous satellite problems and solutions
5. Kepler’s law problems and solutions problems and solutions

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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?

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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1. Newton’s law of universal gravitation problems and solutions
2. Gravitational force, weight problems, and solutions
3. Acceleration due to gravity problems and solutions
4. Geosynchronous satellite problems and solutions
5. Kepler’s law problems and solutions problems and solutions

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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 (m_E) = 5.98 x 10²⁴ kg

Object’s mass (m_o) = 1000 kg

The radius of the Earth (r_E) = 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, m_E = Earth’s mass, m_o = 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

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 :

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 (m_A) = 2

Mass of planet B (m_B) = 3

Radius of the planet A (r_A) = 1

Radius of planet B (r_B) = 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.

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.

1. Newton’s law of universal gravitation problems and solutions
2. Gravitational force, weight problems, and solutions
3. Acceleration due to gravity problems and solutions
4. Geosynchronous satellite problems and solutions
5. Kepler’s law problems and solutions problems and solutions

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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 (d_ob’),

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

d) the distance between the lenses

Known :

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

The focal length of ocular lens (f_oc) = 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 = f_ob / f_oc

M = 400 cm / 20 cm = 20X

The image is virtual and inverted.

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

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

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

If the eye is relaxed, the distance of real image and ocular lens (s_oc) = the focal length of ocular lens (f_oc) = 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 (f_ob) + the focal length of ocular lens (f_oc) = 400 cm + 20 cm = 420 cm.

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 (f_oc) and the focal length of objective lens (f_ob)

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 (f_oc) + the focal length of objective lens (f_ob) = the telescope length (l)

f_oc + f_ob = l

f_ob = l – f_oc

f_ob = 20 – f_oc —–> equation 1

The overall magnification of telescope (M) :

M = f_ob / f_oc

f_ob = M f_oc

f_ob = 40 f_oc —–> equation 2

The focal length of ocular lens :

Subtitute f_ob in equation 1 with f_ob in equation 2 :

f_ob = 20 – f_oc

40 f_oc = 20 – f_oc

40 f_oc + f_oc = 20

41 f_oc = 20

f_oc = 20 cm / 41

f_oc = 0.5 cm

The focal length of ocular lens = 0.5 cm.

The focal length of the objective lens :

f_ob = 20 – f_oc

f_ob = 20 – 0.5

f_ob = 19.5 cm

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2. Convex mirror problems and solutions
3. Diverging lens problems and solutions
4. Converging lens problems and solutions
5. Optical instrument human eye problems and solutions
6. Optical instrument contact lenses problems and solutions
7. Optical instrument eyeglasses
8. Optical instrument magnifying glass problems and solutions
9. Optical instrument microscope – problems and solutions
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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 (f_ob) = 2 cm

The focal length of ocular lens (f_oc) = 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 = m_ob M_oc

M = the overall magnification, m_ob = the magnification of the objective lens, M_oc = the ocular angular magnification

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

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

d_ob’ = l – f_oc = 30 cm – 5 cm = 25 cm

The object distance from the objective lens (d_ob) :

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 (M_ok) :

M_oc = N / f_oc = 25 cm / 5 cm = 5X

The overall magnification :

M = m_ob M_oc = (12.5)(5) = 62.5X

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

The object distance from the objective lens is 2 cm.

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 (m_ob) = 5X

The magnification of ocular lens (M_oc) = 20X

Near point (N) = 25 cm

The distance between the lenses (l) = 15 cm

Solution :

(a) The overall magnification (M)

M = m_ob M_oc

M = (5)(20) = 100X

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

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

M_oc = N / f_oc

The focal length of the ocular lens :

f_oc = N / M_oc = 25 cm / 20 = 1.25 cm

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

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

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

d_ob’ = l – f_oc = 15 cm – 1.25 cm = 13.75 cm

The object distance from the objective lens (d_ob) :

The focal length of the objective lens (f_ob) :

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 (f_ob) = 0.9 cm

The focal length of the ocular lens (f_ok) = 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 (s_ob)

Solution :

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

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

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 (f_ob) = 1.8 cm

The focal length of the ocular lens (f_ok) = 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 (s_ob)

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 (f_ok) + the distance between the image and the objective lens (s_ob’).

s_ob’ = l – f_ok = 24 cm – 6 cm = 18 cm

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

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

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1. Concave mirror problems and solutions
2. Convex mirror problems and solutions
3. Diverging lens problems and solutions
4. Converging lens problems and solutions
5. Optical instrument human eye problems and solutions
6. Optical instrument contact lenses problems and solutions
7. Optical instrument eyeglasses
8. Optical instrument magnifying glass problems and solutions
9. Optical instrument microscope – problems and solutions
10. Optical instrument telescopes problems and solutions

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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 (h_o) = 2 mm

Near point (N) = 25 cm

Object distance (d_o) = 10 cm

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

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

The plus sign indicates that the lens is a converging lens.

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 (h_o) = 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 (h_o) = 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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1. Concave mirror problems and solutions
2. Convex mirror problems and solutions
3. Diverging lens problems and solutions
4. Converging lens problems and solutions
5. Optical instrument human eye problems and solutions
6. Optical instrument contact lenses problems and solutions
7. Optical instrument eyeglasses
8. Optical instrument magnifying glass problems and solutions
9. Optical instrument microscope – problems and solutions
10. Optical instrument telescopes problems and solutions

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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/d_i = 1/f – 1/d_o

-1/d_i = -1/50 – 1/~ = -1/50 – 0

-1/d_i = -1/50

d_i = 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.

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 ?

(c) Nearsighted eye or farsighted 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/d_i = 1/f – 1/d_o

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

–d_i= 575/-2 = -287.5 cm = -2.875 meters

d_i = 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.

The correct answer is B.

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

Solution :

Farsighted eye	Farsighted eye + converging lens

Converging lens = convex lens = positive lens

The correct answer is B.

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

The correct answer is C.

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1. Concave mirror problems and solutions
2. Convex mirror problems and solutions
3. Diverging lens problems and solutions
4. Converging lens problems and solutions
5. Optical instrument human eye problems and solutions
6. Optical instrument contact lenses problems and solutions
7. Optical instrument eyeglasses
8. Optical instrument magnifying glass problems and solutions
9. Optical instrument microscope – problems and solutions
10. Optical instrument telescopes problems and solutions

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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/d_i = 1/f – 1/d_o

-1/d_i = -1/20 – 1/~ = -1/20 – 0

-1/d_i = -1/20

d_i = 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.

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/d_i = 1/f – 1/d_o

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

–d_i = -1675/42 = -40 cm = -0.40 m

d_i = 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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