1. An h-high object is placed at a distance that smaller than focal length f. Determine the properties of the image.

Solution

According to the figure above, properties of the image are virtual, upright and greater than an object.

2. If an object is placed at the center between the focal point and the concave mirror, the image formated by mirror are :

1. Greater 2 times than object

2. Upright

3. Image distance = focal length

4. Virtual

The correct statement is…..

Solution :

Focal length (f) = 20 cm and object distance (d_{o}) = 10 cm, as figure shown below :

a) Image distance

1/f = 1/d_{o} + 1/d_{i}

1/20 = 1/10 + 1/d_{i}

1/20 – 1/10 = 1/d_{i}

1/20 – 2/20 = 1/d_{i}

-1/20 = 1/d_{i}

d_{i} = -20 cm

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

b) Magnification of image

M = -d_{i} / d_{o} = -(-20)/10 = 20/10 = 2 times

The plus sign indicates that the image is upright and the image is greater 2 times than object.

Properties of image

1. Image 2 times greater than the object

2. Upright

3. Image distance = focal length = 20 cm

4. Virtual

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3. An object’s distance is 2 meters in front of the convex mirror and the image height is 1/16 times the object height. What is the focal length of the mirror?

__Known :__

Object distance (d_{o}) = 2 meters

Magnification of image (M) = 1/16 times

__Wanted:__ focal length of the convex mirror

__Solution :__

Image distance :

*The image distance is **– 1/8 meter. **Minus sign indicates that the image is virtual.*

The focal length (f) :

*The minus sign indicates that the focal length is virtual.*

4. Determine the properties of a h-high image, located in front of a convex mirror.

Solution

According to the figure, the properties of the image are virtual, upright and smaller than an object.

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7. A 5-m focal length lens used as a magnifying glass. The lens used by a normal eye when the eye is focused at its near point. What is the angular magnification of the magnifying glass?

__Known :__

The focal length (f) = 5 cm

Near point of normal eye (N) = 25 cm

__Wanted :__ Angular magnification

__Solution :__

If eye is focused at its near point, the image distance = near point of normal eye. The equation of the angular magnification :

*M = Angular magnification, N = near point of normal eye, f = focal length of the magnifying glass*

The angular magnification when the eye is focused at its near point :

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**Nearsighted eye and Farsighted eye**

8. Someone with Nearsighted eye use +2 Diopters eyeglass. What is the closest distance the person can see without eyeglass.

__Known :__

Lens power (P) = +2 diopters

__Wanted :__ The closest distance

__Solution :__

A converging lens or diverging lens?

The power of the lens is positive because the lens used is a positive lens or converging lens.

What is the focal length of the lens?

P = 1/f

2 = 1/f

f = 1/ 2 = 0.5 m = 50 cm

The focal length is 50 cm.

**Nearsighted eye and Farsighted eye?**

Nearsighted eye because the lens used is a positive lens.

**What is the closest distance the person can see without eyeglass?**

In order for the eye to see objects at a distance of 25 cm as a normal eye, the lens should form an image at a distance of x cm in front of the lens. The image is in front of the lens so that the image is upright and virtual. The image is virtual so that the distance of the shadow (s’) is negative.

-1/d_{i }= 1/f – 1/d_{o}

-1/d_{i } = 1/50 – 1/25 = 1/50-2/50 = -1/50

-d_{i }= -50/1 = -50 cm = -0.50 meters

d_{i }= 50 cm = 0.5 meters

The closest distance the person can see without eyeglass is 50 cm. The closest distance a normal eye can see is 25 cm.

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9. The near point of a person with the nearsighted eye is 50 cm. What is the power of lens used by the person so that can see objects at 25 cm?

Solution

Near point of a normal eye is 25 cm and near the point of a person with the nearsighted eye is 50 cm. The lens used by the person is converging lens.

In order for the observed objects to be 25 cm in front of the eye, the lens should form a image at a distance of 50 cm in front of the eye and lens. Image must be in front of the eye to be seen so that the image is upright and virtual.

__Known :__

Object distance (d_{o}) = 25 cm

Image distance (d_{i}) = -50 cm (negative because virtual)

__Wanted ____:__ The focal length (f) and the power of lens (P)

__Solution :__

1/f = 1/s + 1/s’

1/f = 1/25 + 1/-50

1/f = 2/50 – 1/50

1/f = 1/50

f = 50/1 = 50 cm = 0.5 m

Plus sign indicates that the lens is converging.

P = 1/f = 1/0.5 = +2 Diopters

The power of the lens is +2 D.

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10. A microscope has an objective and ocular lens with focal length is 0.9 cm and 5 cm. Someone puts a 10 mm object in front of the objective lens to be observed through an ocular lens without accommodation. If the object has a length of 0.5 mm and the normal reading distance of the person is 25 cm, then the length of the object will be seen to be …

__Known :__

The focal length of the objective lens (f_{ob}) = 0.9 cm = 9 mm

The focal length of the ocular lens (f_{ok}) = 5 cm = 50 mm

Object distance from objective lens (s_{ob}) = 10 mm

Object length (h_{o}) = 0.5 mm

Near point of normal eye (N) = 25 cm = 250 mm

__Wanted :__ Image length (h_{i’})

__Solution :__

**Image distance from objective lens**

1/s_{ob}‘ = 1/f_{ob }– 1/s_{ob} = 1/9 – 1/10 = 10/90 – 9/90 = 1/90

s_{ob}‘ = 90/1 = 90 mm

Total magnification of microscope when the eye is relaxed or when the image at infinity, calculated using equation :

Image length = object length x total magnification = (0.5 mm)(45) = 22.5 mm.

11. The focal length of an objective lens of a microscope is 2.0 cm. An object is placed at 2.2 cm. The length of microscope is 24.5 cm. The eye is normal and relaxed. What is the total magnification of the microscope.

__Known :__

The focal length of the objective lens (f_{ob}) = 2.0 cm

Object distance from objective lens (d_{ob}) = 2.2 cm

Microscope length (l) = 24.5 cm

Near point of normal eye (N) = 25 cm

The eye is relaxed.

__Wanted:__ Total magnification of the microscope (M)

__Solution :__

The eye is relaxed when the image at infinity.

Total magnification of microscope when eye is relaxed :

*M = total magnification*

*l = **microscope length **=**distance between objective lens and ocular lens **= **focal length of ocular lens** + **image distance from objective lens **(d*_{ob}*‘) = f*_{ok }*+ **d*_{ob}*‘ *

__Image distance from objective lens ____(d___{ob}__‘) :__

1/f_{ob} = 1/d_{ob} + 1/d_{ob}‘

1/d_{ob}‘ = 1/f_{ob} – 1/d_{ob} = 1/2 – 1 / 2.2 = 2.2 / 4.4 – 2 / 4.4 = 0.2 / 4.4

d_{ob}‘ = 22

*Image distance from objective lens is 22 cm.*

__Focal length of ocular lens ____(f___{ok}__) :__

l = *f*_{ok }*+ ** **s*_{ob}*‘ *

*f*_{ok }*= l – d*_{ob}*‘ = 24.5 – 22 = 2.5 cm*

__Total magnification of microscope (M) :__

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12. Near point of a normal eye is 25 cm. What is the magnification of the microscope.

__Known :__

Near point of normal eye (N) = 25 cm

Object distance (d_{ob}) = 1.2 cm

Focal length of objective lens (f_{ob}) = 1 cm

Focal length of ocular lens (f_{ok}) = 5 cm

Distance between the objective lens and ocular lens = 10 cm

__Wanted:__ Magnification of microscope

__Solution :__

If the final image at infinity then the eye is relaxed (minimum accommodation). If the final image at near point (25 cm), the eye is not relaxed but focused at near point (maximum accommodation).

Equation of microscope’s magnification when eye is focused at near point (maximum accommodation) :

Image distance from objective lens (d_{ob}‘) :

1/d_{ob}‘ = 1/d_{ob} – 1/d_{ob} = 1/1 – 1 / 1.2 = 12/12 – 10/12 = 2/12

d_{ob}‘ = 12/2 = 6 cm

Image magnification :

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