Verified Solution • 2025–26 Session moderate

QJEEM20Ph06aa: JEE Main 2020 Physics - Class 12 JEEM20Ph06aa Solution: If we need a magnification of 375 from a compound microscope of …

If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece, should be close to :

Multiple Choice Options

  • A 22 mm
  • B 12 mm
  • C 33 mm
  • D 2 mm

Step-by-Step Verified Solution

Case 1 : Near – point adjustment

M.P = Lf0(1+Dfe){L \over {{f_0}}}\left( {1 + {D \over {{f_e}}}} \right)

⇒\Rightarrow 375 = 1505(1+250fe){{150} \over 5}\left( {1 + {{250} \over {{f_e}}}} \right)

⇒\Rightarrow fe = 21.7 mm ≈\approx 22 mm

Case-2 : If final image is at inifinity

M.P = Lf0(Dfe){L \over {{f_0}}}\left( {{D \over {{f_e}}}} \right)

⇒\Rightarrow 375 = 1505(250fe){{150} \over 5}\left( {{{250} \over {{f_e}}}} \right)

⇒\Rightarrow fe = 20 mm

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring