Question

A compound microscope consists of an objective lens of focal length2.0 cm and an eyepiece of focal length 6.25 cm separated by adistance of 15cm. How far from the objective should an object beplaced in order to obtain the final image at (a) the least distance ofdistinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Answer 1

Given that the focal length of the objective lens in a compound microscope is 2.0 cm, focal length of the eyepiece is 15 cm, the separation between the objective lens and eyepiece is 15 cm while the least distance of distinct vision is 25 cm.

(a).

Let f e be the focal length of the eyepiece f o be the focal length of the objective lens, d l be the least distance of the distinct vision, v 2 be the distance of the image formed for the eyepiece and u 2 be the distance of the object from the eyepiece.

From the lens formula,

1 v 2 − 1 u 2 = 1 f e 1 u 2 = 1 v 2 − 1 f e

Substitute the values in the above expression.

1 u 2 = 1 −25 − 1 6.25 = −1−4 25 = −5 25 u 2 =−5 cm

Let v 1 be the distance of the image formed, then

v 1 = d l + u 2

Substitute the values in the above expression.

v 1 =15−5 =10 cm

Let u 1 be the object distance for the objective lens, then by lens formula,

1 v 1 − 1 u 1 = 1 f o 1 u 1 = 1 v 1 − 1 f o

Substitute the values in the above expression.

1 u 1 = 1 10 − 1 2 = 1−5 10 = −4 10 u 1 =−2.5 cm

Hence, the object distance should be −2.5 cm.

Let m be the magnifying power of the microscope, then

m= v 1 | u 1 | ( 1+ d l f e )

Substitute the values in the above expression.

m= 10 2.5 ( 1+ 25 6.25 ) =20

Thus, the value of magnifying power is 20.

(b).

Let v 2 ′ be the image distance for eyepiece, then

v 2 ′ =∞

Use lens formula,

1 v 2 ′ − 1 u 2 ′ = 1 f e 1 u 2 ′ = 1 v 2 ′ − 1 f e

Here, u 2 ′ is the object distance for the eyepiece.

Substitute the values in the above expression.

1 u 2 ′ = 1 ∞ − 1 6.25 u 2 ′ =−6.25

Let v 1 ′ be the image distance for the objective lens, then

v 1 ′ = d l + u 2 ′

Substitute the values in the above expression.

v 1 ′ =15−6.25 =8.75 cm

Let u 1 ′ be the object distance for the objective lens, then by lens formula,

1 v 1 ′ − 1 u 1 ′ = 1 f o 1 u 1 ′ = 1 v 1 ′ − 1 f o

Substitute the values in the above expression.

1 u 1 ′ = 1 8.75 − 1 2 =− 17.5 6.75 =−2.59 cm

Hence, the object distance should be −2.59 cm.

Let m ′ be the magnifying power of the microscope, then

m ′ = v 1 ′ | u 1 ′ | ( d l u 2 ′ )

Substitute the values in the above expression.

m ′ = 8.75 2.59 ( 25 6.25 ) =13.51

Thus, the value of magnifying power is 13.51.