Question

Calculate the magnification of a compound light
microscope with an
objective lenes of focal length-0.6cm and eye pience with a focal length of 3.5cm Seperation between the foci of the two lenses
is 10cm and image is viewed at a distance of 20cm
​

Answer 1

Answer:

length of the objective lens (f

1

)=2.0cm

Focal length of the eyepiece (f

2

)=6.25cm

Distance between the objective lens and the eyepiece (d)=15cm

(a) Least distance of distinct vision,(d

1

)=25cm

Therefore, Image distance for the eyepiece (v

2

)=–25cm

Object distance for the eyepiece =u

2

According to the lens formula,

1/v

2

−1/u

2

=1/f

2

1/u

2

=1/v

2

−1/f

2

=1/−25−1/6.25=−1−4/25=−5/25

u

2

=−5cm

Image distance for the objective lens,(v

1

)=−d+u

2

=15−5=10cm

Object distance for the objective lens =u

2

According to the lens formula,

1/v

1

−1/u

1

=1/f

1

1/u

1

=1/v

1

−1/f

1

=1/10−1/2=1−5/10=−4/10

u

1

=−2.5cm

Magnitude of the object distance (u

1

)=2.5cm