Sure! Let's break this problem down step-by-step.
### Step 1: Understanding the given values
We are given:
- The height of the object ([tex]\( h_o \)[/tex]) is [tex]\( 7.0 \)[/tex] cm.
- The height of the image ([tex]\( h_i \)[/tex]) is [tex]\( 2.8 \)[/tex] cm.
- The distance from the pinhole to the image ([tex]\( d_i \)[/tex]) is [tex]\( 6.0 \)[/tex] cm.
### Step 2: Calculating the magnification
The magnification ([tex]\( M \)[/tex]) of a pinhole camera is defined as the ratio of the image height ([tex]\( h_i \)[/tex]) to the object height ([tex]\( h_o \)[/tex]):
[tex]\[ M = \frac{h_i}{h_o} \][/tex]
Substitute the given values:
[tex]\[ M = \frac{2.8 \, \text{cm}}{7.0 \, \text{cm}} \][/tex]
Performing the division:
[tex]\[ M \approx 0.4 \][/tex]
So, the magnification is approximately [tex]\( 0.4 \)[/tex].
### Step 3: Finding the object distance
The magnification formula can also be written in terms of distances:
[tex]\[ M = \frac{d_i}{d_o} \][/tex]
Where [tex]\( d_o \)[/tex] is the object distance, and [tex]\( d_i \)[/tex] is the image distance (distance from the pinhole to the image).
Rearranging the formula to solve for [tex]\( d_o \)[/tex]:
[tex]\[ d_o = \frac{d_i}{M} \][/tex]
Substitute the given values and previously calculated magnification:
[tex]\[ d_o = \frac{6.0 \, \text{cm}}{0.4} \][/tex]
Performing the division:
[tex]\[ d_o = 15.0 \, \text{cm} \][/tex]
So, the object will be found approximately [tex]\( 15.0 \)[/tex] cm from the pinhole.
### Summary
- The magnification ([tex]\( M \)[/tex]) is approximately [tex]\( 0.4 \)[/tex].
- The object distance ([tex]\( d_o \)[/tex]) is approximately [tex]\( 15.0 \)[/tex] cm.
Thus, the object is approximately [tex]\( 15.0 \)[/tex] cm away from the pinhole in the camera and the magnification is [tex]\( 0.4 \)[/tex].
