Answer :
Sure! Let's solve this step-by-step using the lens formula and magnification formula.
### Step 1: Determine the image distance ([tex]\(v\)[/tex])
1. Lens formula:
[tex]\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \][/tex]
- [tex]\( f \)[/tex] is the focal length of the lens.
- [tex]\( v \)[/tex] is the image distance (what we need to find).
- [tex]\( u \)[/tex] is the object distance.
Rearrange this formula to solve for [tex]\( v \)[/tex]:
[tex]\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \][/tex]
[tex]\[ v = \left( \frac{1}{f} + \frac{1}{u} \right)^{-1} \][/tex]
2. Substitute the values:
- [tex]\( f = 15 \text{ cm} \)[/tex]
- [tex]\( u = 30 \text{ cm} \)[/tex]
[tex]\[ \frac{1}{v} = \frac{1}{15} + \frac{1}{30} \][/tex]
Calculate both fractions:
[tex]\( \frac{1}{15} = \frac{2}{30} \)[/tex]
[tex]\[ \frac{1}{v} = \frac{2}{30} + \frac{1}{30} = \frac{3}{30} = \frac{1}{10} \][/tex]
Therefore,
[tex]\( v = 10 \text{ cm} \)[/tex]
### Step 2: Determine the magnification (M)
1. Magnification formula:
[tex]\[ M = -\frac{v}{u} \][/tex]
2. Substitute the values:
- [tex]\( v = 10 \text{ cm} \)[/tex]
- [tex]\( u = 30 \text{ cm} \)[/tex]
[tex]\[ M = -\frac{10}{30} = -\frac{1}{3} \][/tex]
### Step 3: Determine the image height ([tex]\( h' \)[/tex])
1. Image height formula:
[tex]\[ h' = M \times h \][/tex]
2. Substitute the values:
- [tex]\( M = -\frac{1}{3} \)[/tex]
- [tex]\( h = 10 \text{ cm} \)[/tex]
[tex]\[ h' = -\frac{1}{3} \times 10 = -\frac{10}{3} = -3.33 \text{ cm} \][/tex]
### Step 4: Determine the properties of the image
1. Orientation:
- Since [tex]\( M \)[/tex] is negative, the image is inverted.
2. Size:
- Since [tex]\( |M| = \frac{1}{3} < 1 \)[/tex], the image is diminished.
3. Type:
- Since [tex]\( v \)[/tex] is positive, the image is real.
### Summary
- Image distance ([tex]\( v \)[/tex]) = 10 cm
- Magnification ([tex]\( M \)[/tex]) = -1/3
- Image height ([tex]\( h' \)[/tex]) = -3.33 cm
- Image orientation: Inverted
- Image size: Diminished
- Image type: Real
### Drawing
To draw the ray diagram for the converging lens, follow these steps:
1. Principal Ray: Draw a ray parallel to the principal axis from the top of the object. This ray will refract through the lens and pass through the focal point on the other side.
2. Central Ray: Draw a ray from the top of the object straight through the center of the lens. This ray will continue in a straight line without bending.
3. The point where these rays intersect on the other side of the lens is the top of the inverted image.
The distance from the lens to this intersecting point is the image distance ([tex]\( v \)[/tex]), and the vertical distance from the principal axis at this point is the image height ([tex]\( h' \)[/tex]).
This setup confirms the calculations and illustrates the nature of the image formed by a converging lens.
### Step 1: Determine the image distance ([tex]\(v\)[/tex])
1. Lens formula:
[tex]\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \][/tex]
- [tex]\( f \)[/tex] is the focal length of the lens.
- [tex]\( v \)[/tex] is the image distance (what we need to find).
- [tex]\( u \)[/tex] is the object distance.
Rearrange this formula to solve for [tex]\( v \)[/tex]:
[tex]\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \][/tex]
[tex]\[ v = \left( \frac{1}{f} + \frac{1}{u} \right)^{-1} \][/tex]
2. Substitute the values:
- [tex]\( f = 15 \text{ cm} \)[/tex]
- [tex]\( u = 30 \text{ cm} \)[/tex]
[tex]\[ \frac{1}{v} = \frac{1}{15} + \frac{1}{30} \][/tex]
Calculate both fractions:
[tex]\( \frac{1}{15} = \frac{2}{30} \)[/tex]
[tex]\[ \frac{1}{v} = \frac{2}{30} + \frac{1}{30} = \frac{3}{30} = \frac{1}{10} \][/tex]
Therefore,
[tex]\( v = 10 \text{ cm} \)[/tex]
### Step 2: Determine the magnification (M)
1. Magnification formula:
[tex]\[ M = -\frac{v}{u} \][/tex]
2. Substitute the values:
- [tex]\( v = 10 \text{ cm} \)[/tex]
- [tex]\( u = 30 \text{ cm} \)[/tex]
[tex]\[ M = -\frac{10}{30} = -\frac{1}{3} \][/tex]
### Step 3: Determine the image height ([tex]\( h' \)[/tex])
1. Image height formula:
[tex]\[ h' = M \times h \][/tex]
2. Substitute the values:
- [tex]\( M = -\frac{1}{3} \)[/tex]
- [tex]\( h = 10 \text{ cm} \)[/tex]
[tex]\[ h' = -\frac{1}{3} \times 10 = -\frac{10}{3} = -3.33 \text{ cm} \][/tex]
### Step 4: Determine the properties of the image
1. Orientation:
- Since [tex]\( M \)[/tex] is negative, the image is inverted.
2. Size:
- Since [tex]\( |M| = \frac{1}{3} < 1 \)[/tex], the image is diminished.
3. Type:
- Since [tex]\( v \)[/tex] is positive, the image is real.
### Summary
- Image distance ([tex]\( v \)[/tex]) = 10 cm
- Magnification ([tex]\( M \)[/tex]) = -1/3
- Image height ([tex]\( h' \)[/tex]) = -3.33 cm
- Image orientation: Inverted
- Image size: Diminished
- Image type: Real
### Drawing
To draw the ray diagram for the converging lens, follow these steps:
1. Principal Ray: Draw a ray parallel to the principal axis from the top of the object. This ray will refract through the lens and pass through the focal point on the other side.
2. Central Ray: Draw a ray from the top of the object straight through the center of the lens. This ray will continue in a straight line without bending.
3. The point where these rays intersect on the other side of the lens is the top of the inverted image.
The distance from the lens to this intersecting point is the image distance ([tex]\( v \)[/tex]), and the vertical distance from the principal axis at this point is the image height ([tex]\( h' \)[/tex]).
This setup confirms the calculations and illustrates the nature of the image formed by a converging lens.
