Ngā taputapu whatu karāhe whakanui – ngā raruraru me ngā otinga

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

Mōhiotia:

Object height (h_o) = 2 mm

Near point (N) = 25 cm

ahanoa tawhiti (d_o) = 10 henemita

E hiahiatia ana: Angular magnification (M) and image height (h_i)

Rongoā:

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 kanohi is focused at its near point N = 25 cm (b) angular magnification when the eye is relaxed.

Mōhiotia:

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.

Rongoā:

(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 kanohi is focused at its near point N = 25 cm (b) The image height when the eye is relaxed.

Mōhiotia:

The object height (h_o) = 1 henemita

The focal length (f) = 10 cm

Near point (N) = 25 cm

Rongoā:

(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 ?

Mōhiotia:

Object height (h_o) = 2 mm

Angular magnification (M) = 5X

Near point (N) = 25 cm

Hiahia: Te roa o te arotahi

Rongoā:

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.

Mōhiotia:

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)

E hiahiatia ana: the maximum magnification

Rongoā:

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 wa

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.

Mōhiotia:

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

Power of the magnifying glass (P) = 20 diopters

Hiahia: The minimum magnification

Rongoā:

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

[wpdm_package id='872′]

1. Ngā raruraru me ngā otinga o te whakaata kōpiko
2. Ngā raruraru me ngā otinga o te whakaata pūkohu
3. Ngā raruraru me ngā otinga o te arotahi wehewehe
4. Ngā raruraru me ngā otinga o ngā karāhe arotahi
5. Ngā taputapu whatu Ngā raruraru kanohi tangata me ngā otinga
6. Ngā raruraru me ngā otinga mō ngā karaihe whakapā o ngā taputapu whatu
7. Ngā mōhiti taputapu whatu
8. Ngā raruraru me ngā otinga mō ngā taputapu whatu mō te karāhe whakanui
9. Te karu hikohiko taputapu whatu – ngā raruraru me ngā otinga
10. Ngā raruraru me ngā otinga o ngā karu tirohanga whatu