Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Q 9.12 A person with a normal near point (25 cm) using a compound microscope with an objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope

Answers (1)

best_answer

Inside a microscope,

For the eyepiece lens,

$\frac{1}{f_{eyepiece}}=\frac{1}{v_{eyepiece}} - \frac{1}{u_{eyepiece}}$

we are given

$v_{eyepiece}= -25cm$

$f_{eyepiece}= 2.5cm$

$\frac{1}{2.5}=\frac{1}{-25} - \frac{1}{u_{eyepiece}}$

$\frac{1}{u_{eyepiece}}=\frac{1}{-25} - \frac{1}{2.5}=-\frac{11}{25}$

$u_{eyepiece}=- \frac{25}{11}=-2.27cm$

we can also find this value by finding image distance in the objective lens.

So, in the objective lens

$\frac{1}{f_{objective}}=\frac{1}{v_{objective}} - \frac{1}{u_{objective}}$

we are given

$f_{objective}= 0.8$

$u_{objective}= -0.9$

$\frac{1}{0.8}=\frac{1}{v_{objective}} - \frac{1}{-0.9}$

$\frac{1}{v_{objective}} = \frac{0.1}{0.72}$

$v_{objective} = 7.2cm$

Distance between object lens and eyepiece = $|u_{eyepiece}| + v_{objective}$ = 2.27 + 7.2 = 9.47 cm.

Now,

Magnifying power :

$m = \frac{v_{obejective}}{|u_{objective}|}(1+\frac{d}{f_{eyepiece}})$

$m = \frac{7.2}{0.9}(1+\frac{25}{2.5})= 88$

Hence magnifying power for this case will be 88.

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th

anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

anam-khan

CBSE Class 12 accountancy board exam analysis 2024 out; paper rated easy, balanced

Latest Question