2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Ray Optics

JEE Main & Advanced Physics Ray Optics Question Bank Self Evaluation Test - Ray Optics and Optical Instruments

  • question_answer
    The focal length of the objective and the eyepiece of a telescope are 50 cm and 5 cm respectively. If the telescope is focused for distinct vision on a scale distant 2m from its objective, then its magnifying power will be:

    A) - 4                               

    B) - 8       

    C) +8                               

    D) - 2   

    Correct Answer: D

    Solution :

    [d] Given: \[{{f}_{0}}=50cm,{{f}_{e}}=5cm\] \[d=25cm,{{v}_{0}}=-200cm\] Magnification M=? As \[\frac{1}{{{v}_{0}}}-\frac{1}{{{u}_{0}}}=\frac{1}{{{f}_{\text{0}}}}\] \[\Rightarrow \frac{1}{{{v}_{0}}}=\frac{1}{{{f}_{0}}}+\frac{1}{{{u}_{0}}}=\frac{1}{50}-\frac{1}{200}=\frac{4-1}{200}=\frac{3}{200}\] \[\text{or }{{v}_{0}}=\frac{200}{3}\text{      Now }{{v}_{e}}=d=-25cm\] From, \[\frac{1}{{{v}_{e}}}-\frac{1}{{{u}_{e}}}=\frac{1}{{{f}_{e}}}\] \[-\frac{1}{{{u}_{e}}}=\frac{1}{{{f}_{e}}}-\frac{1}{{{v}_{e}}}=\frac{1}{5}+\frac{1}{25}=\frac{6}{25}\text{ or, }{{v}_{e}}=\frac{-25}{6}cm\]


You need to login to perform this action.
You will be redirected in 3 sec spinner