An eye can distinguish between two points of an object if they are separated by more than 0.22 mm when the object is placed at 25 cm from the eye. The object is now seen by a compound microscope having a 20 D objective and 10 D eyepiece separated by a distance of 20 cm. The final image is formed at 25 cm from the eye. What is the minimum separation between two points of the object which can now be distinguished?
Answers
Given :
Power of the objective lens = 20 D
The focal length of the objective lens is given by
P=1/f
f0=1/20 D=0.05 m=5 cm
Power of the eyepiece = 10 D
The focal length of the eyepiece is given by
fe=1/10 D=0.1 m=10 cm
Least distance for clear vision, D = 25 cm,
To distinguish the two points having minimum separation, the magnifying power should be maximum.
For the eyepiece, we have:
Image distance, ve = − 25 cm
Focal length, fe = 10 cm
The lens formula is given by
1/ue=1/ve-1/fe
=1/-25-1/10
= -2-5/50
=-750
Separation between the objective and the eyepiece = 20 cm
So, the image distance for the objective lens:
v0=20-50/7=90/7 cm
The lens formula for the objective lens is given by
1/u0=1/v0-1/f0
=7/90-1/5=7-18/90
= -11/90
⇒ u0=-90/11 cm
Magnification of the compound microscope:
m=v0/u0[1+D/fe]
=- {[90/7] / -90/11 ] }[ 1+25/10]
=11/7×3.5=5.5
∴ Minimum separation that the eye can distinguish = 0.22/5.5mm = 0.04 mm
An eye can distinguish between two points of an object if they are separated by more than 0.22 mm when the object is placed at 25 cm from the eye. The object is now seen by a compound microscope having a 20 D objective and 10 D eyepiece separated by a distance of 20 cm. The final image is formed at 25 cm from the eye. What is the minimum separation between two points of the object which can now be distinguished?
Solution
. For the given compound microscope. F base 0 = 1/20D = 0.05 m = 5 cm, f base e = 1/10D = 0.1 m = 10 cm. D = 25 cm, separation between objective & eyepiece= 20 cm For the minimum separation between two points which can be distinguished by eye using the microscope, the magnifying power should be maximum. For the eyepiece, v base 0 = –25 cm, f base e = 10 cm So, 1/u base e = 1/v base e – 1/f base e = 1/-25 – 1/10 = - [2 + 5/50] ⇒ u base e = - 50/7 cm So, the image distance for the objective lens should be, V base 0 = 20 – 50/7 = 90/7 cm Now, for the objective lens, 1/u base 0 = 1/v base = - 1/f base 0 = 7/90 – 1/5 = - 11/90 ⇒ u base 0 = - 90/11 cm So, the maximum magnifying power is given by, m = - v base 0/u base 0[1 + D/f base e] = (90/7)/(- 90/11)[1 + 25/10] = 11/7 * 3.5 = 5.5 Thus, minimum separation eye can distinguish = 0.22/5.5 mm = 0.04 mm