Respuesta :
To determine the image distance we will use the following formula:
[tex]\frac{1}{d_i}+\frac{1}{d_0}=\frac{1}{f}[/tex]Where:
[tex]\begin{gathered} d_i=\text{ distance of the image} \\ d_0=\text{ distance of the object} \\ f=\text{ focal distance} \end{gathered}[/tex]We wll solve fo r the imae idistance. We will subtract both sides by 1/d0:
[tex]\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_0}[/tex]Now, we substitute the values:
[tex]\frac{1}{d_i}=\frac{1}{7}-\frac{1}{10}[/tex]Solving the operations:
[tex]\frac{1}{d_i}=\frac{3}{70}[/tex]Now we invert both sides:
[tex]d_i=\frac{70}{3}=23.33cm[/tex]Therefore, the distance of the image is 23.3 cm-.
To determine the image size w us e the followingdformula:
[tex]\frac{d_i}{d_0}=-\frac{h_i}{h_0}[/tex]Where:
[tex]h_i,h_0=\text{ height of the image and height of the object}[/tex]Now, we solve for the height of the image by multiplying both sides by the height of the object:
[tex]-h_0\frac{d_i}{d_0}=h_i[/tex]Now, we substitute the values:
[tex]-(5cm)\frac{23.33cm}{10cm}=h_i[/tex]Solving the operation:
[tex]-11.66cm=h_i[/tex]Therefore, the height ofthe image is -11.66 cm.
The magnification is given by:
[tex]M=\frac{h_i}{h_0}[/tex]Substituting we get:
[tex]M=\frac{23.33cm}{10cm}=2.33[/tex]Therefore, te magnification is 2.33
Since the image height i negative this means that the image isinvdeted.Since the magnification is greater than 1 this means that he image is enlargd. The iamgeimage is real.
