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Pavel [41]
3 years ago
9

In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le

ngth is 25 cm. What is the magnitude of the overall magnification of the microscope?
Physics
1 answer:
expeople1 [14]3 years ago
6 0

Answer:

The  value  is   m \approx   310

Explanation:

From the question we are told that

     The  focal length of the objective is  f_o =  1.0 \ cm

    The  focal length of the eyepiece is  f_e  =  2.0 \  cm

    The  tube length is  L  =  25 \  cm

Generally the magnitude of the overall magnification is mathematically represented as

            m =  m_o  *  m_e

Where  m_o is the objective magnification which is mathematically represented as

        m_o  =  \frac{L}{f_o }

=>      m_o  =  \frac{25}{1 }

=>      m_o  =  25

m_e is the eyepiece magnification which is mathematically evaluated as

     m_e  =  \frac{L }{f_e }

     m_e  =  \frac{25 }{ 2}

      m_e  =  12.5 \  cm

So

    m =  25 * 12.5

     m \approx   310

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Answer:

x(t) = - 6 cos 2t

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Now Again, by Hook's law

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ignoring -ve sign k= 240 N/m

Put given data in eq (1)

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60d²x/dt² + 240x=0

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-2A sin 2(0)+2Bcos(0) =-6

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3 years ago
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