Question

19-iv
The object for the eyepiece would be-
a. The image I, acts as an object for the eye piece.
b. The image 12 acts as an object for the eye piece.
c. The object
d. None of the above​

Answer 1

Explanation:

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