Question

Free Printabl <Week 9 Problem 25.29 - Enhanced - with Solution Part B An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens has a focal length of 12 mm, the eyepiece has a focal length of 28 mm For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of A simple magnifier What is the magnification, including the correct sign of the objective image? IVO AXD ? 130 Submit Previous Answers Request Answer * Incorrect; Try Again; 4 attempts remaining Part C How tall is the image formed by the objective lens? Express your answer with the appropriate units. vi - 13 mm

Answer 1

given
orginal object size, y1 = 1.1 mm
fo = 12 mm
fe = 28 mm

B) object distance for objective lens, uo = 12 + 1

= 13 mm

let ve is the image distance.

use, 1/uo + 1/vo = 1/fo

1/13 + 1/vo = 1/12

1/vo = 1/12 - 1/13

vo = 156 mm

magnification, m1 = -vo/uo

= -156/13

= -12 <<<<<<<<<-------------Answer

C) image height, y1' = m*object height

= -12*1.1

= -13.2 mm <<<<<<<<<-------------Answer

D) Angular magnification eye piece, M = D/fe (here D is near point of a normal person)

= 25/2.8

= 8.93 <<<<<<<<<-------------Answer

part F) M = m1*M

= -12*8.93

= -107.2 <<<<<<<<<-------------Answer

part G) M = (25 cm)*s1'/(f1*f2)

= (25 cm)*vo/(f1*f2)

= 25*15.6/(1.1*2.8)

= 126.6 <<<<<<<<<-------------Answer