Respuesta :
Answer:
The focal length of the lens and the height of the image are -6.86 cm and -0.638 cm.
Explanation:
Given that,
Distance of the object u = -16.0 cm
Distance of the image v = -12.0 cm
Height of the object = 8.50 mm = 0.85 cm
(a). We need to calculate the focal length
Using lens's formula
[tex]\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}[/tex]
Where, f = focal length
u = distance of the object
v = distance of the image
Put the value into the formula
[tex]\dfrac{1}{f}=\dfrac{1}{-16.0}+\dfrac{1}{-12.0}[/tex]
[tex]\dfrac{1}{f}=-\dfrac{7}{48}[/tex]
[tex]f=-\dfrac{48}{7}[/tex]
[tex]f=-6.86\ cm[/tex]
The lens is diverging.
(b). Using formula of magnification
[tex]m=-\dfrac{v}{u}[/tex]
[tex]-\dfrac{v}{u}=\dfrac{h'}{h}[/tex]
Put the value in to the formula
[tex]-\dfrac{12.0}{16.0}=\dfrac{h'}{0.85}[/tex]
[tex]h'=-\dfrac{12.0\times0.85}{16.0}[/tex]
[tex]h'=-0.638\ cm[/tex]
The image is inverted.
Hence, The focal length of the lens and the height of the image are -6.86 cm and -0.638 cm.
