A compound microscope has lenses of focal length 1.0cm and 5.0cm. An object is placed 1.2cm from the objective lens. If a virtual image is formed 25cm from the eyepiece, calculate the distance between the two lenses
Answer
1) Using the lens formula for image q formed by the objective.
"\\frac{1}{f_o} =\\frac{1}{p}+\\frac{1}{q}\\\\\n\n\\frac{1}{1}=\\frac{1}{1.2}+\\frac{1}{q}\\\\\n\nq=6 cm."
2) Now, this image will act as an object for second lens i.e. eyepiece. Using the lens formula for the distance p_1 from the eyepiece.
"\\frac{1}{f_e} =\\frac{1}{p_1 }+\\frac{1}{q_1}\\\\\n\n\\frac{1}{5}=\\frac{1}{p_1} +\\frac{1}{(-25)}\\\\\n\np_1=4.2 cm."
3) The distance between the two lenses:
"L=6+4.2=10.2 cm."
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