Respuesta :
Answer:
Image is virtual and formed on the same side as the object. 2 m from the lens.
The size of the image is 7.97 cm
Image is upright as the magnification is positive and smaller than the object.
Explanation:
u = Object distance = 6 m
v = Image distance
f = Focal length = -3 m (concave lens)
[tex]h_u[/tex]= Object height = 24 cm
Lens Equation
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{-3}-\frac{1}{6}\\\Rightarrow \frac{1}{v}=\frac{-1}{2} \\\Rightarrow v=\frac{-2}{1}=-2\ m[/tex]
Image is virtual and formed on the same side as the object. 2 m from the lens.
Magnification
[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-2}{6}\\\Rightarrow m=0.33[/tex]
[tex]m=\frac{h_v}{h_u}\\\Rightarrow 0.33=\frac{h_v}{0.24}\\\Rightarrow h_v=0.33\times 0.24=0.0797\ m[/tex]
The size of the image is 7.97 cm
Image is upright as the magnification is positive and smaller than the object.
