If in compound microscope objective with focal length 1.0 cm, eyepiece with focal length 2.0 cm, tube length of 20 cm and near point for an observer is 25 cm. Then value of magnification of compound microscope will be .

Question

Accepted Answer

(3) 250. Given: objective fₒ=1.0 cm, eyepiece fₑ=2.0 cm, tube length L=20 cm, near point D=25 cm. Idea: the objective forms a real image near the far end of the tube, magnified about L/(fₒ) times; the eyepiece then acts as a simple magnifier of angular magnification D/(fₑ) when the eye is relaxed. The two multiply: m=L/(fₒ)×D/(fₑ). Objective: (20)/(1.0)=20. Eyepiece: (25)/(2.0)=12.5. So m=20×12.5=250. (With the final image at the near point instead, the eyepiece factor would be 1+D/(fₑ)=13.5 and m=270. That value is not among the options, so the paper means the relaxed eye, with the image at infinity.)

If in compound microscope objective with focal length 1.0 cm, eyepiece with focal length 2.0 cm, tube length of 20 cm and near point for an observer is 25 cm. Then value of magnification of compound microscope will be ________.

Step-by-step solution

Why the other options are wrong