Answer :
To determine the focal length of a concave lens given the object distance and the image size, we can use the lens formula and magnification concept. Follow these steps:
1. Identify the given data:
- Object distance ([tex]\(d_o\)[/tex]): 20 cm
- The virtual image is three times smaller than the object, so the magnification ([tex]\(m\)[/tex]): [tex]\(-\frac{1}{3}\)[/tex]
2. Determine the image distance ([tex]\(d_i\)[/tex]):
The magnification formula relates the image distance to the object distance and is given by:
[tex]\[ m = \frac{d_i}{d_o} \][/tex]
Given [tex]\(m = -\frac{1}{3}\)[/tex], [tex]\(-\frac{1}{3} = \frac{d_i}{20}\)[/tex].
3. Solve for [tex]\(d_i\)[/tex]:
[tex]\[ d_i = -\frac{1}{3} \times 20 = -\frac{20}{3} \text{ cm} \][/tex]
The negative sign indicates that the image is virtual and formed on the same side as the object.
4. Use the lens formula to find the focal length ([tex]\(f\)[/tex]):
The lens formula is:
[tex]\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \][/tex]
Substitute [tex]\(d_o = 20 \text{ cm}\)[/tex] and [tex]\(d_i = -\frac{20}{3} \text{ cm}\)[/tex] into the formula:
[tex]\[ \frac{1}{f} = \frac{1}{20} + \frac{1}{-\frac{20}{3}} \][/tex]
5. Simplify the expression:
[tex]\[ \frac{1}{f} = \frac{1}{20} - \frac{3}{20} \][/tex]
[tex]\[ \frac{1}{f} = \frac{1 - 3}{20} = \frac{-2}{20} = -\frac{1}{10} \][/tex]
6. Solve for [tex]\(f\)[/tex]:
[tex]\[ f = -10 \text{ cm} \][/tex]
Therefore, the focal length of the concave lens is [tex]\(-10 \text{ cm}\)[/tex]. The negative sign indicates that it is a concave lens.
1. Identify the given data:
- Object distance ([tex]\(d_o\)[/tex]): 20 cm
- The virtual image is three times smaller than the object, so the magnification ([tex]\(m\)[/tex]): [tex]\(-\frac{1}{3}\)[/tex]
2. Determine the image distance ([tex]\(d_i\)[/tex]):
The magnification formula relates the image distance to the object distance and is given by:
[tex]\[ m = \frac{d_i}{d_o} \][/tex]
Given [tex]\(m = -\frac{1}{3}\)[/tex], [tex]\(-\frac{1}{3} = \frac{d_i}{20}\)[/tex].
3. Solve for [tex]\(d_i\)[/tex]:
[tex]\[ d_i = -\frac{1}{3} \times 20 = -\frac{20}{3} \text{ cm} \][/tex]
The negative sign indicates that the image is virtual and formed on the same side as the object.
4. Use the lens formula to find the focal length ([tex]\(f\)[/tex]):
The lens formula is:
[tex]\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \][/tex]
Substitute [tex]\(d_o = 20 \text{ cm}\)[/tex] and [tex]\(d_i = -\frac{20}{3} \text{ cm}\)[/tex] into the formula:
[tex]\[ \frac{1}{f} = \frac{1}{20} + \frac{1}{-\frac{20}{3}} \][/tex]
5. Simplify the expression:
[tex]\[ \frac{1}{f} = \frac{1}{20} - \frac{3}{20} \][/tex]
[tex]\[ \frac{1}{f} = \frac{1 - 3}{20} = \frac{-2}{20} = -\frac{1}{10} \][/tex]
6. Solve for [tex]\(f\)[/tex]:
[tex]\[ f = -10 \text{ cm} \][/tex]
Therefore, the focal length of the concave lens is [tex]\(-10 \text{ cm}\)[/tex]. The negative sign indicates that it is a concave lens.
