Physics, asked by mounibeauty2082, 12 days ago

An object is placed at a distance 1cm from objective of compound microscope.Distance between lenses is 30cm and the intermediate image at is 5cm distance from eyepiece find magnifying power.

Answers

Answered by llchummill
0

Answer:

Focal length of the objective lens f

1

=2cm

Focal length of the eyepiece,f

2

=6.25cm

Distance between the objective lens and the eyepiece,d=15cm

Least distance of distinct vision,d

=25cm

Image distance for the eyepiece,v

2

=−25cm

bject distance for the eyepiece = u

2

According to the lens formula

f

2

1

=

v

2

1

u

2

1

6.25

1

=

−25

1

u

2

1

u

2

=−5cm

Image distance for the objective lens v

1

=d+u

2

=15−5=10cm

According to the lens formula

f

1

1

=

v

1

1

u

1

1

2

1

=

10

1

u

1

1

u

2

=−2.5cm

The magnifying power of a compound microscope is given by the relation

m=

∣u

1

v

1

(1+

f

2

d

)

m=

2.5

10

(1+

6.25

25

)=20

b)The final image is formed at infinity

Image distance for the eyepiece v

2

=∞

According to the lens formula

f

2

1

=

v

2

1

u

2

1

According to the lens formula

6.25

1

=

1

u

2

1

u

2

=−6.25cm

Image distance for the objective lens v

1

=d+u

2

=10−6.25=8.75

Object distance for the objective len =u

2

According to the lens formula

f

1

1

=

v

1

1

u

1

1

2

1

=

8.75

1

u

1

1

u

2

=−2.59cm

The magnifying power of a compound microscope is given by the relation:

∣u

1

v

1

(

∣u

2

d

)=13.51

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