# 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/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.

**Ebook PDF Optical instrument eyeglasses sample problems with solutions 46.71 KB**

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