Answer
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Hint: This problem can be solved with the help of basic knowledge of magnification in the case of microscopes. One must remember the formula for magnification produced by the eyepiece of the compound microscope in order to solve this question. also, the formula for magnification produced by the microscope must also be known.
Formula Used: We will use the following formula to solve this question
\[{m_e} = 1 + \dfrac{D}{{{f_e}}}\]
And
\[m = {m_o} \times {m_e}\]
Where
\[{m_e}\] is the magnification of the eyepiece
\[D\] is the least distance of distinct vision
\[{f_e}\] is the focal length of the eyepiece
\[m\] is the magnifying power
\[{m_o}\] is the magnification of the objective
Complete Step-by-Step Solution:
The following information is provided to us in the question:
The magnification power of the compound microscope is \[30\]
The focal length of the eyepiece, \[{f_e} = 5 cm\]
The magnification of the eyepiece is given by
\[{m_e} = 1 + \dfrac{D}{{{f_e}}}\]
Upon substituting the values, we have
\[{m_e} = 1 + \dfrac{{25}}{5} = 6\]
Now, we will put the value of magnification of eyepiece into the second formula listed above
That is
\[m = {m_o} \times {m_e}\]
Upon substituting values, we get
\[30 = {m_o} \times 6\]
On further solving,
\[\therefore {m_o} = 5\]
Hence, the magnification produced by the objective is \[5\].
Additional Information: Using a combination of lenses, real and magnified images of minuscule particles or objects can be achieved. A compound microscope is an intricate collection of a mixture of lenses that gives a highly maximized and magnified picture of living microscopic entities and other complex details or cells and tissues.
Note: We know that viewing objects that are so small that they are invisible to the eye is the main use of the microscope. There are various applications of this device depending on the fields it is used. It can be used to solve crimes, to cure illnesses, to produce new materials and even to analyse fossils that have been part of history.
Formula Used: We will use the following formula to solve this question
\[{m_e} = 1 + \dfrac{D}{{{f_e}}}\]
And
\[m = {m_o} \times {m_e}\]
Where
\[{m_e}\] is the magnification of the eyepiece
\[D\] is the least distance of distinct vision
\[{f_e}\] is the focal length of the eyepiece
\[m\] is the magnifying power
\[{m_o}\] is the magnification of the objective
Complete Step-by-Step Solution:
The following information is provided to us in the question:
The magnification power of the compound microscope is \[30\]
The focal length of the eyepiece, \[{f_e} = 5 cm\]
The magnification of the eyepiece is given by
\[{m_e} = 1 + \dfrac{D}{{{f_e}}}\]
Upon substituting the values, we have
\[{m_e} = 1 + \dfrac{{25}}{5} = 6\]
Now, we will put the value of magnification of eyepiece into the second formula listed above
That is
\[m = {m_o} \times {m_e}\]
Upon substituting values, we get
\[30 = {m_o} \times 6\]
On further solving,
\[\therefore {m_o} = 5\]
Hence, the magnification produced by the objective is \[5\].
Additional Information: Using a combination of lenses, real and magnified images of minuscule particles or objects can be achieved. A compound microscope is an intricate collection of a mixture of lenses that gives a highly maximized and magnified picture of living microscopic entities and other complex details or cells and tissues.
Note: We know that viewing objects that are so small that they are invisible to the eye is the main use of the microscope. There are various applications of this device depending on the fields it is used. It can be used to solve crimes, to cure illnesses, to produce new materials and even to analyse fossils that have been part of history.
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