# A Compound Microscope Consists of an Objective Lens of Focal Length 2.0 Cm and an Eyepiece of Focal Length 6.25 Cm Separated by a Distance of 15 Cm. - Physics

Numerical

A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

#### Solution

Focal length of the objective lens, f1 = 2.0 cm

Focal length of the eyepiece, f2 = 6.25 cm

Distance between the objective lens and the eyepiece, d = 15 cm

(a) Least distance of distinct vision, d' = 25

∴ Image distance for the eyepiece, v2 = −25 cm

Object distance for the eyepiece = u2

According to the lens formula, we have the relation:

1/"v"_2 - 1/"u"_2 = 1/"f"_2

1/"u"_2 = 1/"v"_2 - 1/"f"_2

= 1/(-25) - 1/6.25

= (-1 - 14)/25

= (-5)/25

∴ u2 = −5 cm

Image distance for the objective lens, v1 = d + u2 = 15 − 5 = 10 cm

Object distance for the objective lens = u1

According to the lens formula, we have the relation:

1/"v"_1 - 1/"u"_1 = 1/"f"_1

1/"u"_1 = 1/"v"_1 - 1/"f"_1

= 1/10 - 1/2

= (1 - 5)/10

= (-4)/10

∴ u1 = −2.5 cm

Magnitude of the object distance, |u1| = 2.5 cm

The magnifying power of a compound microscope is given by the relation:

"m" = "v"_1/|"u"_1| (1 + "d'"/"f"^2)

= 10/2.5 (1+ 25/6.25)

= 4 × (1 + 4)

= 20

Hence, the magnifying power of the microscope is 20.

(b) The final image is formed at infinity.

∴ Image distance for the eyepiece, v2 = ∞

Object distance for the eyepiece = u2

According to the lens formula, we have the relation:

1/"v"_2 - 1/"u"_2 = 1/"f"_2

1/∞ - 1/"u"_2 = 1/6.25

∴ u2 = −6.25 cm

Image distance for the objective lens, v1 = d + u2 = 15 − 6.25 = 8.75 cm

Object distance for the objective lens = u1

According to the lens formula, we have the relation:

1/"v"_1 - 1/"u"_1 = 1/"f"_1

1/"u"_1 = 1/"v"_1 - 1/"f"_1

= 1/8.75 - 1/2.0

= (2 - 8.75)/17.5

∴ "u"_1 = -17.5/6.75 = −2.59 cm

Magnitude of the object distance, |u1| = 2.59 cm

The magnifying power of a compound microscope is given by the relation:

"m" = "v"_1/|"u"_1| (("d'")/|"u"_2|)

= 8.75/2.59  xx 25/6.25

= 13.51

Hence, the magnifying power of the microscope is 13.51.

Concept: Optical Instruments - The Microscope
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 9 Ray Optics and Optical Instruments
Exercise | Q 9.11 | Page 345
NCERT Class 12 Physics Textbook
Chapter 9 Ray Optics and Optical Instruments
Exercise | Q 11 | Page 346

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