Question

An insect 0.80 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 26 mm

Part A:  Where is the image formed by the objective lens? Give your answer as the distance from the image to the lens. Express your answer in millimeters to two significant figures.

Part B: What is the magnification, including the correct sign, of the objective image? Express your answer to two significant figures.

Part C: How tall is the image formed by the objective lens? Express your answer in millimeters to two significant figures.

Part D: What is the angular magnification of the eyepiece? Express your answer using two significant figures.

Part E: Using your results from parts B and D, calculate the magnitude of the overall magnification of the microscope. Express your answer using two significant figures.

Part F: Use M=m1M2= (25cm)s′1/f1f2 (Eq. 25.4) to calculate the overall magnification of the microscope.

Express your answer using two significant figures.

Part G: Explain why the answers to part E and F differ.

Answer 1

let h = 0.8 mm
for object lens, f1 = 12 mm
object distance, u1 = 12 + 1 = 13 mm

part A) let v1 is the image distance.

use, 1/u1 + 1/v1 = 1/f1

1/v1 = 1/f1 - 1/u1

1/v1 = 1/12 - 1/13

v1 = 156 mm

part B) magnification, m1 = -v1/u1

= -156/13

= -12

part C) image height = m*object height

= -12*0.8

= -9.6 mm (negaitve represents image is inverted)

part D) angular magnification of eye peice, M = 250/fe

= 250/26

= 9.615

part E) magnitude of overall magniifcation = m1*M

= 12*9.615

= 115

part F) M = m1*M2

= 25 cm*v1/(f1*f2)

= (250*156)/(12*26)

= 125

G)part E represents magnification when object is formed at near point.

paart F represents magnification when final image is formed at infinite point.