1.

The focal length of the objective and the eye piece of a compound microscope are 2.0cm and 3.0 respectively. The distance between the objective and the eyepiece is 15.0cm, The final image formed by the eyepiece is at infinity. The two lenses are thin. The distance, in cm, of the object and the image produced by the objective, measured from the objective lens, are respectively,

Answer» <html><body><p></p>Solution :The eyepiece forms the final image at infinity.Its object should therefore lie at its <a href="https://interviewquestions.tuteehub.com/tag/focus-25840" style="font-weight:bold;" target="_blank" title="Click to know more about FOCUS">FOCUS</a>. F denotes focus of eyepiece. I denotes image formed by the <a href="https://interviewquestions.tuteehub.com/tag/objective-25531" style="font-weight:bold;" target="_blank" title="Click to know more about OBJECTIVE">OBJECTIVE</a> lens which serves as object for eyepiece. IT should be at <a href="https://interviewquestions.tuteehub.com/tag/3cm-310901" style="font-weight:bold;" target="_blank" title="Click to know more about 3CM">3CM</a> from eyepiece. <br/> `therefore v_0` for objective lens =15-3=12 cm (<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) <br/> `therefore 1/v_2-1/u_0=1/f_0` or <br/> `1/12-1/mu_0=1/2` <br/> `1/mu_0=1/12-1/2=-5 /12` <br/> `1/mu_0=1/12-1/2= -5/12 ` <br/> or `mu_0=-2.4cm` From objective lens `mu_0=2.4cm `(to left) `v_0=12 cm` (to right)<br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_DOC_OBJ_PHY_XII_V02_A_C02_SLV_113_S01.png" width="80%"/></body></html>


Discussion

No Comment Found

Related InterviewSolutions