Question

17. The two lenses of a compound microscope are of focal lengths 2 cm and 5 cm. If an object is placed
at a distance of 2.1 cm from the objective of focal length 2 cm, the final image forms at the least
distance of distinct vision of a normal eye. Find the magnifying power of the microscope (NCERT
1) 20
2) 6
3) 120
4) 60​

Answer 1

Answer:

هاها شوطًا طويلًا يجب أن تكون خطيرًا علينا في قوتك إلى السلطة قريبًا ، فالسلطة والسيطرة لا تقومان بأفضل طريقة لعام 111123

Answer 2

The magnifying power of microscope is 12.0 cm

Explanation:

Given that,

Focal length of first lens = 2 cm

Focal length of eye piece = 5 cm

Object distance u= 2.1 cm

Least distance of distinct vision D= 25 cm

Image distance for the eye piece v'= -25 cm

We need to calculate the image distance for second lens

Using formula of lens

$\dfrac{1}{f'}=\dfrac{1}{v'}-\dfrac{1}{u'}$

Put the value into the formula

$\dfrac{1}{5}=\dfrac{1}{-25}-\dfrac{1}{u'}$

$-\dfrac{1}{u'}=\dfrac{1}{5}+\dfrac{1}{25}$

$-\dfrac{1}{u'}=\dfrac{6}{25}$

$u'=-4.2\ cm$

We need to calculate the image distance for the objective lens

Using formula of image distance

$v=-d+u'$

Put the value into the formula

$v=2.1-4.2$

$v=-2.1\ cm$

We need to calculate the object distance for first lens

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{2}=\dfrac{1}{-2.1}-\dfrac{1}{u}$

$-\dfrac{1}{u}=\dfrac{1}{2}+\dfrac{1}{2.1}$

$u=-\dfrac{41}{42}$

$u=-1.024\ cm$

We need to calculate the magnifying power of microscope

Using formula of magnification

$m=\dfrac{v}{u}(1+\dfrac{D}{f_{e})$

$m=\dfrac{2.1}{1.024}(1+\dfrac{25}{5})$

$m=12.00$

Hence, The magnifying power of microscope is 12.0

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Topic ; optics

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