Respuesta :
Answer:
The focal length is -0.5.
Explanation:
Given that,
Magnification [tex]m=\dfrac{1}{2}[/tex]
Object distance u = 0.5 cm
We need to calculate the image distance
Using formula of magnification
[tex]m=-\dfrac{v}{u}[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}=-\dfrac{v}{0.5}[/tex]
[tex]v=-\dfrac{1}{2}\times0.5[/tex]
[tex]v=-0.25\ cm[/tex]
We need to calculate the focal length
Using formula for focal length
[tex]\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}[/tex]
[tex]\dfrac{1}{f}=\dfrac{v+u}{vu}[/tex]
[tex]f=\dfrac{uv}{v+u}[/tex]
[tex]f=\dfrac{0.5\times(-0.25)}{-0.25+0.5}[/tex]
[tex]f=-0.5\ cm[/tex]
Hence, The focal length is -0.5.
Answer:
[tex]- 0.5[/tex] cm
Explanation:
[tex]M[/tex] = Lateral Magnification = 0.5
[tex]d_{o}[/tex] = Distance of the object = 0.5 cm
[tex]d_{i}[/tex] = Distance of the image = ?
Lateral magnification is given as
[tex]M = \frac{- d_{i}}{d_{o}}[/tex]
[tex]0.5 = \frac{- d_{i}}{0.5}[/tex]
[tex]d_{i} = - 0.25[/tex] cm
[tex]f[/tex] = focal length of the magnifier
Using the equation
[tex]\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}[/tex]
[tex]\frac{1}{0.5} + \frac{- 1}{0.25} = \frac{1}{f}[/tex]
[tex]f = - 0.5[/tex] cm
