Physics Circle home Student Login

Home›Chapters›Ray Optics PYQs›Optical Instruments & the Eye›Question 44

JEE Main 2014 (11 Apr, Shift 1) · Optical Instruments & the Eye

Optical Instruments & the Eye question from JEE Main 2014 (11 Apr, Shift 1)

Q44JEE Main 2014 · 11 APR · SHIFT 1Medium

Optical Instruments & the Eye

In a compound microscope the focal length of objective lens is 1.2 cm and focal length of eye piece is 3.0 cm. When object is kept at 1.25 cm in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be:

Show answer and solution

Answer: (a)

  1. Objective: 1vo=11.2−11.25=0.051.5=130⇒vo=30\dfrac{1}{v_o} = \dfrac{1}{1.2} - \dfrac{1}{1.25} = \dfrac{0.05}{1.5} = \dfrac{1}{30} \Rightarrow v_o = 30 cm.
  2. mo=vouo=301.25=24m_o = \dfrac{v_o}{u_o} = \dfrac{30}{1.25} = 24.
  3. Image at infinity: me=Dfe=253m_e = \dfrac{D}{f_e} = \dfrac{25}{3}.
  4. M=24×253=200M = 24 \times \dfrac{25}{3} = 200.
  5. So the answer is (a) 200.
Report a mistake

More questions on Optical Instruments & the Eye