Answer to Question #261509 in Optics for Nbj

Question

Q 2. In a compound microscope, the focal length of the objective lens is 5 cm. 1 mm tall

object is placed 7 cm in front of the objective lens and the eyepiece is focused so that an

inverted image is formed 8 cm in front of the objective lens. The distance between the

objective lens and the eyepiece is 30 cm.

(Hint: Draw a rough diagram and indicate the length measurements as described above)

(a) Determine the focal length of the eyepiece

(b) What it the total magnification of the image

Answer 1

${Focal\,length\,of\,the\,objective\,lens\,f_1 = 5.1cm}\\ {Focal\,length\,of\,the\,eyepiece\,lens\,f_2}\\ {object\,distance\,of\,the\,objective\,lens\,u_1 = 7.0cm}\\ {object\,distance\,of\,the\,eyepiece\,lens\,u_2}\\ {image\,distance\,of\,the\,objective\,lens\,v_1 }\\ {image\,distance\,of\,the\,eyepiece\,lens\,v_2 = 38.0cm}\\ {The\,distance\,between\,the\,eyepiece\,and\,objective\,d = 30.0cm}\\ {Q2\,(a)}\\ {According\,to\,the\,lens\,formula}\\ \frac{1}{f_1} = \frac{1}{v_1}-\frac{1}{u_1}\\ \frac{1}{5.1} = \frac{1}{v_1}-\frac{1}{7.0}\\ \frac{1}{v_1} = \frac{1}{5.1}-\frac{1}{7.0}\\ \frac{1}{v_1} = \frac{121}{357}\\ {v_1}={2.95cm}\\ \\ {object\,distance\,of\,the\,eyepiece\,lens\,u_2}\\ {u_2 = v_1-d}\\ {u_2 = 2.95-30}\\ {u_2 = -27.1cm}\\\\ {According\,to\,the\,lens\,formula}\\ \frac{1}{f_2} = \frac{1}{v_2}-\frac{1}{u_2}\\ \frac{1}{f_2} = \frac{1}{-38}-\frac{1}{-27.1}\\ \frac{1}{f_2} = \frac{109}{10298}\\ {f_2} = {94.5cm}\\ {2}{b}\\ {m = \frac{v_1}{u_1}{(1+\frac{d'}{f_2})}}\\ {m = \frac{3.0}{7.0}{(1+\frac{38}{94.5})}}\\ {m = {\times0.6}}$