Question

A compound microscope has an objective of focal length 1 cm and an eyepiece of 4 cm. If the tube length is 20 cm. Then the magnification of the compound microscope is : (final image is at D )

• A. 145
• B. 14.5
• C. 1.45
• D. None of these

Answer 1

The correct option is A 145

Answer is A.
Magnification produced by the objective is given by:
Mo= size of image / size of object
That is, Mo= q1 / p1
In order to get maximum magnification, we must decrease p1 and increase q1. Thus maximum possible value of p1 is fo i.e., p1 = fo and maximum possible value of q1 is the length of microscope i.e., q1 = L
So, Mo = L / fe - eqn 1
Magnification produced by the eye piece is given by:
Me= size of image / size of object
Me= q2 / p2
We know that the eye piece behaves as a magnifying glass therefore the final image will be formed at a least distance of distinct vision i.e at 25 cm from the eye. Hence q2 = d.
Me = d / p2 - eqn 2
Using thin lens formula for eye piece :
1/f2 = 1/q2 + 1/p2
Here f2 = fe, q2 = - d and p = p2
1/fe = 1/-d + 1/p2
1/fe = -1/d + 1/p2
Multiplying both sides by "d"
d/fe = -d/d + d/p2
d/fe = -1 + d/p2
1 + d/fe = d/p2
d/p2 = 1 + d/fe - eqn 3
Comparing equation (2) and (3)
Me = 1 + d/fe
Therefore, Me = 1 + (25 / fe) - eqn 4
Total magnification is equal to the product of the magnification produced by the objective and the eye piece.
M = Mo * Me
M = (L/fo)(1 + 25/fe)
In this case, L = 20 cm, fo = 1 cm and fe = 4 cm.
So, M = (20/1)(1+25/4) = 145.
Hence, the magnification of the compound microscope is 145.