Respuesta :
Answer:
[tex]d_{o}[/tex] = 154 cm
[tex]d_{i}[/tex] = 216.6 cm
The image is real
Explanation:
[tex]h_{o}[/tex] = height of the object = 3.20 cm
[tex]h_{i}[/tex] = height of the image = 4.50 cm
f = focal length of the converging lens = 90 cm
[tex]d_{o}[/tex] = object distance from the lens = ?
[tex]d_{i}[/tex] = image distance from the lens = ?
using the equation for magnification
[tex]\frac{h_{i}}{h_{o}}= \frac{ d_{i}}{d_{o}}[/tex]
[tex]\frac{4.50}{3.20}= \frac{d_{i}}{d_{o}}[/tex]
[tex]d_{i}[/tex] = 1.40625 [tex]d_{o}[/tex] eq-1
using the lens equation
[tex]\frac{1}{d_{i}} + \frac{1}{d_{o}} = \frac{1}{f}[/tex]
using eq-1
[tex]\frac{1}{( 1.40625)d_{o}} + \frac{1}{d_{o}} = \frac{1}{90}[/tex]
[tex]d_{o}[/tex] = 154 cm
Using eq-1
[tex]d_{i}[/tex] = 1.40625 [tex]d_{o}[/tex]
[tex]d_{i}[/tex] = 1.40625 (154)
[tex]d_{i}[/tex] = 216.6 cm
The image is real
