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Draw a ray diagram to show the working of a compound microscope. Deduce an expression for the total magnification when the final image is formed at the near point.

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It is an instrument used to see highly magnified image of tiny objects.

Principle: It is based on the principle that when a small object is placed just outside the focus of objective lens, its real, inverted and magnified image is formed on the other side of the lens. This image acts as an object for eye lens, the eye lens is so adjusted that the final image is produced at least distance of distinct vision.

Construction:. It consists of two convex lenses of shorter focal lengths arranged co-axially at the ends of two sliding tubes.

(i) Objective: It is a convex lens of very short focal length and of small aperture and is placed near the object to be magnified.

(ii) Eye piece: It is a convex lens of short focal length and of comparatively large aperture and is placed near the eye for seeing the final image which is formed at least distance of distinct vision.

Eye piece

The distance between the two lenses can be adjusted by rack and pinion arrangement.

Let a tiny object AB be placed in front of objective at a distance more than f0. Its real and enlarged image is formed at A' B'. This image A'B' acts as an object for eye piece and forms final image at A"B" i.e. at a distance D, the distance of distinct vision.

Magnifying Power of Microscope is defined as the ratio of the angle subtended on the eye by the final image to the angle subtended on the eye by the object, when both the final image and the object are situated at least distance of distinct vision.

Magnifying power, M = \(\frac{\beta}{\alpha}\)

Since angles are small, therefore, tan ß ≃ ß and tan α ≃ α.

Magnifying power

where me and m0 are the magnifying powers of eye piece and the objective.

m0\(\frac{A'B}{AB} = \frac{CB'}{CB}\)

But CB is nearly equal to (-f0) and CB' is equal to the length of the tube L [since the image A'B' is formed very close to eye lens]

m0\(\frac{L}{(-f_0)}\)

Also me = 1 + \(\frac{D}{f_e}\) [As in simple microscope]

∴ M = \(-\frac{L}{f_0} ( 1 + \frac{D}{f_e})\)

(Negative sign shows that final image is inverted with respect to the object.)

Thus we find that, if the focal length of objective lens and eye lens are short, the magnifying power will be large. Magnifying power of compound microscope when final image is at infinity. Magnifying power is defined as the angle subtended on the eye by the final image to the angle subtanded on the eye by the object, at the unaided eye, when final image is formed at infinity. When final image is formed at infinity.

focal length

The angular magnification due to eye piece me = \(\frac{D}{f_e}\)

The angular magnification due to objective lens m0\(\frac{L}{f_0}\)

∴ M = m0me = \(\frac{L}{f_0} \times \frac{D}{f_e}\) 

Thus we find that, to have large magnification, the focal length of the objective lens and eye lens must be small.

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