Question

A compound microscope has an objective of focal length 1.25cm and eyepic of focal length 5cm object is kept at 2.5 cm from objective the final image is formed at infinity find the distance between the objective and eye piece

Answer 1

Answer:

5.7cm

Explanation:

according to your question

focal length of objective =1.25cm

focal length of eyepiece =5cm

object distance =2.5cm

image distance =infinite

for objective lens :

1/v-1/u=1/f

=>1/v=1/f+1/u

=>1/v=1/1.25+1/2.5

=>1/v=100/125+10/25

=>1/v=6/5

=>v=6/5cm

this is the image distance for objective lens

but we have given that the image distance for the eyepiece is infinite

.

so the object of eyepiece if present at the focus point

so the distance between the objective and eyepiece is

=5/6+5

=35/6

=5.7cm

I think this will help you

Answer 2

Dear Student,

◆ Answer -

L = 7.5 cm

● Explanation -

# Given -

fo = +1.25 cm

fe = +5 cm

uo = -2.5 cm

ve = ∞

# Solution -

Applying lens formula for objective,

1/fo = 1/vo - 1/uo

1/1.25 = 1/vo - 1/-2.5

1/vo = 1/25 - 1/2.5

1/vo = 1/2.5

vo = 2.5 cm

Applying lens formula for eyepiece,

1/fe = 1/ve - 1/ue

1/5 = 1/∞ - 1/ue

1/5 = 0 - 1/ue

ue = -5 cm

Distance between the objective and eyepiece is -

L = |vo| + |ue|

L = |2.5| + |-5|

L = 2.5 + 5

L = 7.5 cm

Hence, distance between the objective and eyepiece is 7.5 cm.

Thanks dear. Hope this helps you..