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