Explanation:
Given that,
Height of object = 4.0 cm
Distance of the object u= -30.0 cm
Focal length = 23 cm
We need to calculate the image distance
Using lens's formula
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}[/tex]
Where, u = object distance
v = image distance
f = focal length
Put the value into the formula
[tex]\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}[/tex]
[tex]\dfrac{1}{v}=\dfrac{7}{690}[/tex]
[tex]v=98.57\ cm[/tex]
We need to calculate the height of the image
Using formula of height
[tex]\dfrac{h'}{h}=\dfrac{-v}{u}[/tex]
Where, h' = height of image
h = height of object
[tex]\dfrac{h'}{4}=\dfrac{-98.57}{-30}[/tex]
[tex]h'=\dfrac{98.57\times4}{30}[/tex]
[tex]h'=13.14\ cm[/tex]
The image is real and inverted.
(b). Now, object distance u = 13.0 cm
We need to calculate the image distance
Using lens's formula
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}[/tex]
Put the value into the formula
[tex]\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}[/tex]
[tex]\dfrac{1}{v}=-\dfrac{10}{299}[/tex]
[tex]v=-29.9\ cm[/tex]
We need to calculate the height of the image
[tex]\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}[/tex]
[tex]h=-\dfrac{29.9\times4}{13.0}[/tex]
[tex]h=-9.2\ cm[/tex]
The image is virtual and erect.
Hence, This is required solution.