JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
In a compound microscope, the magnified virtual image is formed at a distance of \(25\, cm\) from the eye-piece. The focal length of its objective lens is \(1\, cm .\) If the magnification is \(100\) and the tube length of the microscope is \(20\, cm ,\) then the focal length of the eye-piece lens (in \(cm\) ) is
- A \(4.5\)
- B \(4\)
- C \(3.5\)
- D \(4.8\)
Answer & Solution
Correct Answer
(A) \(4.5\)
Step-by-step Solution
Detailed explanation
for first lens \(=\frac{1}{v_{1}}-\frac{1}{-x}=\frac{1}{1} \Rightarrow v_{1}=\frac{x}{x-1}\) also magnification \(\left| m _{1}\right|=\left|\frac{ v _{1}}{ u _{1}}\right|=\frac{1}{ x -1}\) for \(2^{nd}\) lens this is acting as object so…
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