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NCERT Class XII Chapter
Ray Optics and Optical Instruments
Questions With Solutions

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Question : 12
Total: 38
A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Solution:  
The image is formed at least distance of distinct vision for sharp focus. The separation between two lenses will be vo+|ue|

Let us find first vo the image distance for objective lens.
1
v0
−
1
u0
 =
1
f0
 or
1
v0
 -
1
−9
 =
1
8
 or
1
v0
 =
1
8
−
1
9
 =
1
72
⇒ v0 = 72 mm = 7.2 cm
Also we can find object distance for eyepiece ue, as we know
ve = D = 25 cm = 250 mm
1
ve
−
1
ue
 =
1
fe
 ⇒
1
−250
−/ue =
1
25
−
1
ue
 =
1
25
+
1
250
 =
10+1
250
ue = -
250
11
 mm = - 22.7 mm = - 2.27 cm
Separation between lenses
L = vo+|ue| = 72 + 22.7 or L = 94.7 mm = 9.47 cm
Magnifying power m = −
v0
u0
[1+
D
fe
]
m = -
72
−9
[1+
250
25
] = 8 [11] = 88
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