Answer :
Let's solve the problem step-by-step:
### Given Data
- Object height ([tex]\(h_o\)[/tex]): 10 cm
- Object distance ([tex]\(u\)[/tex]): 5 cm in front of the mirror (it will be negative in sign, [tex]\(u = -5\)[/tex] cm)
- Radius of curvature ([tex]\(R\)[/tex]): 40 cm (for a concave mirror, it is positive)
### Formulas Needed
1. Focal length ([tex]\(f\)[/tex]) of the mirror:
[tex]\[ f = \frac{R}{2} \][/tex]
2. Mirror formula:
[tex]\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \][/tex]
Where:
- [tex]\( f \)[/tex] = focal length
- [tex]\( v \)[/tex] = image distance
- [tex]\( u \)[/tex] = object distance
3. Magnification ([tex]\(m\)[/tex]):
[tex]\[ m = - \frac{v}{u} = \frac{h_i}{h_o} \][/tex]
Where:
- [tex]\( m \)[/tex] = magnification
- [tex]\( h_i \)[/tex] = image height
- [tex]\( h_o \)[/tex] = object height
### Solution
#### (I) Determining the position of the image
1. Calculate the focal length ([tex]\(f\)[/tex]):
[tex]\[ f = \frac{R}{2} = \frac{40 \, \text{cm}}{2} = 20 \, \text{cm} \][/tex]
2. Apply the mirror formula to find [tex]\(v\)[/tex] (image distance):
[tex]\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \][/tex]
[tex]\[ \frac{1}{20} = \frac{1}{v} + \frac{1}{-5} \][/tex]
Rearrange and solve for [tex]\(\frac{1}{v}\)[/tex]:
[tex]\[ \frac{1}{v} = \frac{1}{20} + \frac{1}{5} \][/tex]
Convert [tex]\(\frac{1}{5}\)[/tex] to a fraction with the same denominator as [tex]\(\frac{1}{20}\)[/tex]:
[tex]\[ \frac{1}{5} = \frac{4}{20} \][/tex]
Add the fractions:
[tex]\[ \frac{1}{v} = \frac{1}{20} + \frac{4}{20} = \frac{5}{20} = \frac{1}{4} \][/tex]
Therefore, the image distance [tex]\(v\)[/tex] is:
[tex]\[ v = 4 \, \text{cm} \][/tex]
#### (II) Determining the size of the image
1. Calculate the magnification ([tex]\(m\)[/tex]):
[tex]\[ m = - \frac{v}{u} \][/tex]
[tex]\[ m = - \frac{4}{-5} = \frac{4}{5} = 0.8 \][/tex]
2. Using magnification to find the image height ([tex]\(h_i\)[/tex]):
[tex]\[ m = \frac{h_i}{h_o} \][/tex]
[tex]\[ 0.8 = \frac{h_i}{10 \, \text{cm}} \][/tex]
Solve for [tex]\(h_i\)[/tex]:
[tex]\[ h_i = 0.8 \times 10 \, \text{cm} = 8 \, \text{cm} \][/tex]
So, the image height is 8 cm.
#### Summary of Results:
1. Image distance (v): The image is formed at a distance of 4 cm in front of the mirror (since [tex]\(v\)[/tex] is positive, it confirms the image is real and on the same side as the object).
2. Image size (h_i): The size of the image is 8 cm.
#### (III) Drawing a Ray Diagram
Here is a step-by-step guide to drawing the ray diagram:
1. Draw the principal axis: A straight horizontal line.
2. Draw the concave mirror: A curved line with the center at the vertex.
3. Mark the focal point (F): Halfway between the vertex and the center of curvature.
4. Mark the center of curvature (C): Twice the focal length away from the vertex.
5. Place the object: 5 cm in front of the mirror (closer than the focal point).
6. Draw rays:
- A ray parallel to the principal axis from the top of the object refracts through the focal point.
- A ray passing through the focal point before striking the mirror travels parallel to the principal axis after reflection.
- A ray directed towards the center of curvature reflects back onto itself.
7. Locate the intersection: The intersection of these reflected rays gives the position of the image.
The diagram will show an image formed between the focal point and the vertex of the mirror.
In conclusion, the image is real, inverted, and smaller than the object, located between the focal point and the vertex of the concave mirror.
### Given Data
- Object height ([tex]\(h_o\)[/tex]): 10 cm
- Object distance ([tex]\(u\)[/tex]): 5 cm in front of the mirror (it will be negative in sign, [tex]\(u = -5\)[/tex] cm)
- Radius of curvature ([tex]\(R\)[/tex]): 40 cm (for a concave mirror, it is positive)
### Formulas Needed
1. Focal length ([tex]\(f\)[/tex]) of the mirror:
[tex]\[ f = \frac{R}{2} \][/tex]
2. Mirror formula:
[tex]\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \][/tex]
Where:
- [tex]\( f \)[/tex] = focal length
- [tex]\( v \)[/tex] = image distance
- [tex]\( u \)[/tex] = object distance
3. Magnification ([tex]\(m\)[/tex]):
[tex]\[ m = - \frac{v}{u} = \frac{h_i}{h_o} \][/tex]
Where:
- [tex]\( m \)[/tex] = magnification
- [tex]\( h_i \)[/tex] = image height
- [tex]\( h_o \)[/tex] = object height
### Solution
#### (I) Determining the position of the image
1. Calculate the focal length ([tex]\(f\)[/tex]):
[tex]\[ f = \frac{R}{2} = \frac{40 \, \text{cm}}{2} = 20 \, \text{cm} \][/tex]
2. Apply the mirror formula to find [tex]\(v\)[/tex] (image distance):
[tex]\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \][/tex]
[tex]\[ \frac{1}{20} = \frac{1}{v} + \frac{1}{-5} \][/tex]
Rearrange and solve for [tex]\(\frac{1}{v}\)[/tex]:
[tex]\[ \frac{1}{v} = \frac{1}{20} + \frac{1}{5} \][/tex]
Convert [tex]\(\frac{1}{5}\)[/tex] to a fraction with the same denominator as [tex]\(\frac{1}{20}\)[/tex]:
[tex]\[ \frac{1}{5} = \frac{4}{20} \][/tex]
Add the fractions:
[tex]\[ \frac{1}{v} = \frac{1}{20} + \frac{4}{20} = \frac{5}{20} = \frac{1}{4} \][/tex]
Therefore, the image distance [tex]\(v\)[/tex] is:
[tex]\[ v = 4 \, \text{cm} \][/tex]
#### (II) Determining the size of the image
1. Calculate the magnification ([tex]\(m\)[/tex]):
[tex]\[ m = - \frac{v}{u} \][/tex]
[tex]\[ m = - \frac{4}{-5} = \frac{4}{5} = 0.8 \][/tex]
2. Using magnification to find the image height ([tex]\(h_i\)[/tex]):
[tex]\[ m = \frac{h_i}{h_o} \][/tex]
[tex]\[ 0.8 = \frac{h_i}{10 \, \text{cm}} \][/tex]
Solve for [tex]\(h_i\)[/tex]:
[tex]\[ h_i = 0.8 \times 10 \, \text{cm} = 8 \, \text{cm} \][/tex]
So, the image height is 8 cm.
#### Summary of Results:
1. Image distance (v): The image is formed at a distance of 4 cm in front of the mirror (since [tex]\(v\)[/tex] is positive, it confirms the image is real and on the same side as the object).
2. Image size (h_i): The size of the image is 8 cm.
#### (III) Drawing a Ray Diagram
Here is a step-by-step guide to drawing the ray diagram:
1. Draw the principal axis: A straight horizontal line.
2. Draw the concave mirror: A curved line with the center at the vertex.
3. Mark the focal point (F): Halfway between the vertex and the center of curvature.
4. Mark the center of curvature (C): Twice the focal length away from the vertex.
5. Place the object: 5 cm in front of the mirror (closer than the focal point).
6. Draw rays:
- A ray parallel to the principal axis from the top of the object refracts through the focal point.
- A ray passing through the focal point before striking the mirror travels parallel to the principal axis after reflection.
- A ray directed towards the center of curvature reflects back onto itself.
7. Locate the intersection: The intersection of these reflected rays gives the position of the image.
The diagram will show an image formed between the focal point and the vertex of the mirror.
In conclusion, the image is real, inverted, and smaller than the object, located between the focal point and the vertex of the concave mirror.
