KCET · Physics · Dual Nature of Matter
A microscope is having objective of focal length \( 1 \mathrm{~cm} \) and eyepiece of focal length \( 6 \mathrm{~cm} \). If tube
length is \( 30 \mathrm{~cm} \) and image is formed at the least distance of distinct vision, what is the
magnification produced by the microscope ? Take D \( =25 \mathrm{~cm} \).
- A \( 06 \)
- B \( 150 \)
- C 15
- D \( 125 \)
Answer & Solution
Correct Answer
(B) \( 150 \)
Step-by-step Solution
Detailed explanation
Given, focal length of objective, \( f_{o}=1 \mathrm{~cm} \); focal length of eyepiece, \( f_{e}=6 \mathrm{~cm} \); length of tube, \( L=30 \)
Using
\( M=\frac{L}{f_{0}}\left(1+\frac{D}{f_{e}}\right)=\frac{30}{1}\left(1+\frac{25}{6}\right)=30(1+4.166) \) \( \Rightarrow M=155=150 \)
Therefore, magnification produced by the microscope is, approximately, \( 150 \).
Using
\( M=\frac{L}{f_{0}}\left(1+\frac{D}{f_{e}}\right)=\frac{30}{1}\left(1+\frac{25}{6}\right)=30(1+4.166) \) \( \Rightarrow M=155=150 \)
Therefore, magnification produced by the microscope is, approximately, \( 150 \).
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