Question

27. (a) What magnification is produced by a 0.150 cm focal length microscope objective that is 0.155 cm from the object being viewed? (b) What is the overall magnification if an 8× eyepiece (one that produces a magnification of 8.00) is used?

Answer 1

Answer:

magnification = - 30

overall magnification = -240

Explanation:

given data

Focal length of microscope objective f = 0.150 cm

Object distance from microscope objective do = 0.155 cm

magnification by eyepiece = 8 ×

to find out

What magnification is produced  and overall magnification

solution

we consider here  Image distance from microscope objective is =  di

so that

Magnification produced by objective will be = - $\frac{di}{do}$

so we find here  di by given equation that is

$\frac{1}{do} +\frac{1}{di} = \frac{1}{f}$     ..................1

$\frac{1}{di} = \frac{1}{0.150} - \frac{1}{0.155}$

di = 4.65 cm

so that magnification by object will be

magnification = - $\frac{di}{do}$

magnification = - $\frac{4.65}{0.155}$

magnification = - 30

and

overall magnification will be

overall magnification = magnification by objective × magnification by eyepiece   ........................2

overall magnification = -30 × 8

overall magnification = -240