Answer:
The value is [tex]m \approx 310[/tex]
Explanation:
From the question we are told that
The focal length of the objective is [tex]f_o = 1.0 \ cm[/tex]
The focal length of the eyepiece is [tex]f_e = 2.0 \ cm[/tex]
The tube length is [tex]L = 25 \ cm[/tex]
Generally the magnitude of the overall magnification is mathematically represented as
[tex]m = m_o * m_e[/tex]
Where [tex]m_o[/tex] is the objective magnification which is mathematically represented as
[tex]m_o = \frac{L}{f_o }[/tex]
=> [tex]m_o = \frac{25}{1 }[/tex]
=> [tex]m_o = 25[/tex]
[tex]m_e[/tex] is the eyepiece magnification which is mathematically evaluated as
[tex]m_e = \frac{L }{f_e }[/tex]
[tex]m_e = \frac{25 }{ 2}[/tex]
[tex]m_e = 12.5 \ cm[/tex]
So
[tex]m = 25 * 12.5[/tex]
[tex]m \approx 310[/tex]
