Certainly! Let's solve the problem step by step.
### Given Data:
- Length of the flame ([tex]\(h\)[/tex]) = 1.5 cm
- Focal length of the mirror ([tex]\(f\)[/tex]) = 12 cm
- Distance of the flame from the mirror ([tex]\(u\)[/tex]) = 18 cm
### Required:
a) Distance of the image from the mirror ([tex]\(v\)[/tex])
b) Length of the image ([tex]\(h'\)[/tex])
### Step 1: Use the Mirror Formula to Find the Image Distance ([tex]\(v\)[/tex])
The mirror formula for a concave mirror is:
[tex]\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \][/tex]
Rearranging for [tex]\( \frac{1}{v} \)[/tex]:
[tex]\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \][/tex]
Substitute the known values into the formula:
[tex]\[ \frac{1}{v} = \frac{1}{12} - \frac{1}{18} \][/tex]
Find a common denominator to simplify the fractions:
[tex]\[ \frac{1}{v} = \frac{3 - 2}{36} \][/tex]
[tex]\[ \frac{1}{v} = \frac{1}{36} \][/tex]
Therefore, the image distance [tex]\(v\)[/tex] is:
[tex]\[ v = 36 \text{ cm} \][/tex]
### Step 2: Use the Magnification Formula to Find the Image Length ([tex]\(h'\)[/tex])
The magnification ([tex]\(m\)[/tex]) of a mirror is given by:
[tex]\[ m = -\frac{v}{u} \][/tex]
Where [tex]\(m\)[/tex] is also the ratio of the image height ([tex]\(h'\)[/tex]) to the object height ([tex]\(h\)[/tex]), so:
[tex]\[ m = \frac{h'}{h} \][/tex]
Substitute the values of [tex]\(v\)[/tex] and [tex]\(u\)[/tex] into the magnification formula:
[tex]\[ m = -\frac{36}{18} \][/tex]
[tex]\[ m = -2 \][/tex]
So, the magnification ([tex]\(m\)[/tex]) is [tex]\(-2\)[/tex].
Using the magnification to find the image length ([tex]\(h'\)[/tex]):
[tex]\[ h' = m \times h \][/tex]
[tex]\[ h' = -2 \times 1.5 \][/tex]
[tex]\[ h' = -3 \text{ cm} \][/tex]
The negative sign indicates that the image is inverted relative to the object.
### Final Answers:
a) The distance of the image from the mirror is [tex]\(36 \text{ cm} \)[/tex].
b) The length of the image is [tex]\(3 \text{ cm}\)[/tex] (inverted).
