JEE PYQ: Ray Optics Question 25
Question 25 - 2020 (07 Jan Shift 1)
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:
(1) 22 mm
(2) 12 mm
(3) 2 mm
(4) 33 mm
Show Answer
Answer: (1)
Solution
According to question, $M = 375$ $L = 150$ mm, $f_0 = 5$ mm and $f_e = ?$ Using, magnification, $M = \frac{L}{f_0}\left(1 + \frac{D}{f_e}\right)$ $\Rightarrow 375 = \frac{150}{5}\left(1 + \frac{250}{f_e}\right)$ ($\because D = 25$ cm $= 250$ mm) $\Rightarrow 12.5 = 1 + \frac{250}{f_e}$ $\Rightarrow f_e = \frac{250}{11.5} = 21.7 \approx 22$ mm
BETA
