Respuesta :
Answer:
(a). The overall magnification is 175.
(b). The overall magnification is 472.
Explanation:
Given that,
Focal length of lens = 1.40 mm
Focal length of eyepiece = 20.00 mm
Near point = 25.0 cm
Object distance = 1.50 mm
(a). We need to calculate the image distance
Using formula of lens
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}[/tex]
Put the value into the formula
[tex]\dfrac{1}{1.40}=\dfrac{1}{v}+\dfrac{1}{1.50}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{1.40}-\dfrac{1}{1.50}[/tex]
[tex]v=21\ mm[/tex]
We need to calculate the magnification of objective lens
Using formula of magnification
[tex]m=\dfrac{v}{u}[/tex]
Put the value into the formula
[tex]m=\dfrac{21}{1.50}[/tex]
[tex]m=14[/tex]
We need to calculate the length
Using formula for length
[tex]l=f_{e}+m u[/tex]
Put the value into the formula
[tex]l=20.0+14\times1.50[/tex]
[tex]l=41[/tex]
We need to calculate the overall magnification
Using formula of magnification
[tex]M=\dfrac{N}{f_{e}}(\dfrac{l-f_{e}}{u})[/tex]
Put the value into the formula
[tex]M=\dfrac{250.0}{20.0}(\dfrac{41-20.0}{1.50})[/tex]
[tex]M=175[/tex]
The overall magnification is 175.
(b). Focal length = 3.00 mm
Focal length of eyepiece = 6.0
Near point = 25 cm
Length = 40.0 cm
We need to calculate the overall magnification
Using formula of magnification
[tex]M=\dfrac{N}{f_{e}}(\dfrac{l-f_{e}}{u})[/tex]
Put the value into the formula
[tex]M=\dfrac{25}{6.00}\times\dfrac{40-6.00}{0.3}[/tex]
[tex]M=472[/tex]
Hence, (a). The overall magnification is 175.
(b). The overall magnification is 472.
