Certainly! Let's solve this step-by-step.
### Step 1: Understand the Problem
We have:
- An object of height [tex]\( h_o = 7.5 \)[/tex] cm.
- The object is placed [tex]\( u = 35 \)[/tex] cm from a converging lens.
- The focal length of the lens [tex]\( f = 15 \)[/tex] cm.
We need to determine:
1. The image distance ([tex]\( v \)[/tex]).
2. The magnification ([tex]\( M \)[/tex]).
3. The height of the image ([tex]\( h_i \)[/tex]).
### Step 2: Use the Lens Formula to Find the Image Distance
The lens formula is given by:
[tex]\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \][/tex]
We rearrange the formula to solve for [tex]\( v \)[/tex]:
[tex]\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \][/tex]
### Step 3: Plug in the Known Values
First, note that the object distance ([tex]\( u \)[/tex]) is taken as negative for real objects placed in front of the lens:
[tex]\[ u = -35 \text{ cm} \][/tex]
Now substitute the known values into the rearranged lens formula:
[tex]\[ \frac{1}{v} = \frac{1}{15} + \frac{1}{-35} \][/tex]
Calculate:
[tex]\[ \frac{1}{15} \approx 0.0667 \][/tex]
[tex]\[ \frac{1}{-35} \approx -0.0286 \][/tex]
Adding these together gives:
[tex]\[ \frac{1}{v} \approx 0.0667 - 0.0286 = 0.0381 \][/tex]
Thus, the image distance [tex]\( v \)[/tex] is approximately:
[tex]\[ v = \frac{1}{0.0381} \approx 26.25 \text{ cm} \][/tex]
### Step 4: Calculate the Magnification
The magnification ([tex]\( M \)[/tex]) of the lens is given by:
[tex]\[ M = -\frac{v}{u} \][/tex]
Substituting the values:
[tex]\[ M = -\frac{26.25}{-35} \][/tex]
Calculate:
[tex]\[ M = \frac{26.25}{35} \approx 0.75 \][/tex]
### Step 5: Determine the Height of the Image
The height of the image ([tex]\( h_i \)[/tex]) can be found using the magnification:
[tex]\[ h_i = M \times h_o \][/tex]
Substituting the values:
[tex]\[ h_i = 0.75 \times 7.5 \][/tex]
Calculate:
[tex]\[ h_i \approx 5.625 \text{ cm} \][/tex]
### Step 6: Summarize the Results
- The image distance, [tex]\( v \)[/tex], is approximately [tex]\( 26.25 \)[/tex] cm.
- The magnification, [tex]\( M \)[/tex], is [tex]\( 0.75 \)[/tex].
- The height of the image, [tex]\( h_i \)[/tex], is approximately [tex]\( 5.625 \)[/tex] cm.
### Nature of the Image:
- Since the magnification is positive, the image is upright relative to the object.
- The image distance ([tex]\( v \)[/tex]) being positive indicates the image is real and formed on the opposite side of the lens from the object.
So, the image is real, upright, and has a height of about [tex]\( 5.625 \)[/tex] cm with a magnification of [tex]\( 0.75 \)[/tex].
