Answer :
Let's solve this step-by-step, starting with finding the focal length of the mirror and then the height of the image.
1. Given Values:
- Object height (h) = 2 cm
- Object distance (u) = 12 cm (will be taken as negative because it is in front of the mirror)
- Image distance (v) = 4 cm (will also be taken as negative because it is in front of the mirror for a concave mirror)
2. Sign Convention:
According to the sign convention for mirrors:
- Distances measured against the direction of incident light (towards the mirror) are taken as negative.
Hence,
- Object distance (u) = -12 cm
- Image distance (v) = -4 cm
3. Mirror Formula:
The mirror formula relates the object distance (u), image distance (v), and focal length (f) of a mirror:
[tex]\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \][/tex]
Substituting the given values:
[tex]\[ \frac{1}{f} = \frac{1}{-4} + \frac{1}{-12} \][/tex]
Solving the right-hand side:
[tex]\[ \frac{1}{f} = -\frac{1}{4} - \frac{1}{12} \][/tex]
To add these fractions, find a common denominator (12):
[tex]\[ \frac{1}{f} = -\frac{3}{12} - \frac{1}{12} \][/tex]
[tex]\[ \frac{1}{f} = -\frac{4}{12} \][/tex]
[tex]\[ \frac{1}{f} = -\frac{1}{3} \][/tex]
Taking the reciprocal to find the focal length:
[tex]\[ f = -3 \text{ cm} \][/tex]
4. Magnification:
The magnification (m) produced by a mirror is given by:
[tex]\[ m = \frac{\text{Height of the image (h')}}{\text{Height of the object (h)}} = -\frac{v}{u} \][/tex]
Using the given values:
[tex]\[ m = -\frac{-4}{-12} = \frac{4}{12} = \frac{1}{3} \][/tex]
5. Height of the Image:
The height of the image (h') can be found using the magnification formula:
[tex]\[ h' = m \times h = \frac{1}{3} \times 2 \][/tex]
[tex]\[ h' = \frac{2}{3} \text{ cm} \][/tex]
Therefore, the focal length of the mirror is -3 cm, and the height of the image is [tex]\(-\frac{2}{3}\)[/tex] cm (or [tex]\(-0.67\)[/tex] cm) which indicates that the image is inverted.
1. Given Values:
- Object height (h) = 2 cm
- Object distance (u) = 12 cm (will be taken as negative because it is in front of the mirror)
- Image distance (v) = 4 cm (will also be taken as negative because it is in front of the mirror for a concave mirror)
2. Sign Convention:
According to the sign convention for mirrors:
- Distances measured against the direction of incident light (towards the mirror) are taken as negative.
Hence,
- Object distance (u) = -12 cm
- Image distance (v) = -4 cm
3. Mirror Formula:
The mirror formula relates the object distance (u), image distance (v), and focal length (f) of a mirror:
[tex]\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \][/tex]
Substituting the given values:
[tex]\[ \frac{1}{f} = \frac{1}{-4} + \frac{1}{-12} \][/tex]
Solving the right-hand side:
[tex]\[ \frac{1}{f} = -\frac{1}{4} - \frac{1}{12} \][/tex]
To add these fractions, find a common denominator (12):
[tex]\[ \frac{1}{f} = -\frac{3}{12} - \frac{1}{12} \][/tex]
[tex]\[ \frac{1}{f} = -\frac{4}{12} \][/tex]
[tex]\[ \frac{1}{f} = -\frac{1}{3} \][/tex]
Taking the reciprocal to find the focal length:
[tex]\[ f = -3 \text{ cm} \][/tex]
4. Magnification:
The magnification (m) produced by a mirror is given by:
[tex]\[ m = \frac{\text{Height of the image (h')}}{\text{Height of the object (h)}} = -\frac{v}{u} \][/tex]
Using the given values:
[tex]\[ m = -\frac{-4}{-12} = \frac{4}{12} = \frac{1}{3} \][/tex]
5. Height of the Image:
The height of the image (h') can be found using the magnification formula:
[tex]\[ h' = m \times h = \frac{1}{3} \times 2 \][/tex]
[tex]\[ h' = \frac{2}{3} \text{ cm} \][/tex]
Therefore, the focal length of the mirror is -3 cm, and the height of the image is [tex]\(-\frac{2}{3}\)[/tex] cm (or [tex]\(-0.67\)[/tex] cm) which indicates that the image is inverted.
