PhysEduc › Class 12 Physics › Chapter 9: Ray Optics and Optical Instruments

The microscope

NCERT Class 12 Physics, Chapter 9: Ray Optics and Optical Instruments, §9.7.1. A short visual lesson: 2 narrated 3D videos, 2 interactive simulations and 2 quiz questions.

The simple microscope

Narrated video. Transcript:

How big something looks depends on the angle it makes at your eye. Bring an ant closer and it looks bigger, because the angle grows.

But you cannot bring it closer than about 25 centimetres and still see it sharply. This is the near point, D = 25 centimetres. At D, an object of height h makes an angle θ₀ = h/D.

A simple microscope is just a convex lens of short focal length: a magnifying glass. Put the object inside its focus, and the lens forms a virtual, upright, magnified image.

Adjust it so the image is at the near point, v = −D. The magnification is m = v/u. Using the lens formula, m = 1 − v/f, which becomes m = 1 + D/f.

For a lens of focal length 5 centimetres, m = 1 + 25/5 = 6. The object sits at 1/u = −1/25 − 1/5, which gives u ≈ −4.2 centimetres.

Looking at a near image strains the eye. It is more relaxing to put the object exactly at the focus. The image then goes to infinity, and the rays enter the eye parallel.

The image now makes an angle θᵢ = h/f at the eye. The angular magnification is θᵢ over θ₀, which is (h/f) ÷ (h/D) = D/f. For our 5 centimetre lens, that is 5.

A shorter focal length gives more magnification. But a single lens of very short focal length shows distortions, so a simple magnifier cannot go much beyond about ten times.

The compound microscope

Narrated video. Transcript:

For more magnification we use two lenses. The compound microscope has a small objective of short focal length fₒ near the object, and an eyepiece of focal length fₑ near the eye.

The object is placed just beyond the focus of the objective. The objective forms a real, inverted, magnified image A′B′ inside the tube.

The eyepiece then acts as a simple microscope for this image, giving a final virtual, magnified image. Compared with the object, the final image is inverted.

Let L be the tube length: the distance between the second focus of the objective and the first focus of the eyepiece. From similar triangles, tan β = h/fₒ = h′/L, so the objective magnifies by mₒ = L/fₒ.

The eyepiece magnifies by mₑ = 1 + D/fₑ with the final image at the near point, or D/fₑ with it at infinity. The total magnification is m = mₒ × mₑ.

With the final image at infinity, m = (L/fₒ)(D/fₑ). Take fₒ = 1 centimetre, fₑ = 2 centimetres and L = 20 centimetres: m = 20 × 12.5 = 250.

Another example: fₒ = 2 centimetres and fₑ = 6.25 centimetres, 15 centimetres apart. For the final image at 25 centimetres, the eyepiece needs its object at uₑ = −5 centimetres.

So the objective’s image is 15 − 5 = 10 centimetres from it. Then 1/uₒ = 1/10 − 1/2, giving uₒ = −2.5 centimetres. mₒ = 10 ÷ 2.5 = 4, mₑ = 1 + 25/6.25 = 5, and m = 20.

Real microscopes use several lenses in both objective and eyepiece to get sharp images, and can magnify a thousand times or more.

Try it: a magnifying glass

Interactive simulation. Things to try:

  • A 5 cm lens with the object at 4.17 cm: the virtual image is at about 25 cm, and the angular magnification is about 6 = 1 + D/f.
  • Move the object out to the focus (u = −5 cm). The image goes to infinity and the magnification becomes D/f = 5.
  • Try a shorter focal length. The magnification goes up.

Try it: build a compound microscope

Interactive simulation. Things to try:

  • This is the example: fₒ = 2 cm, fₑ = 6.25 cm, 15 cm apart. Check uₒ = −2.5 cm, mₒ = −4, mₑ = 5 and m = 20.
  • The objective forms a real image I₁ just inside the eyepiece’s focus Fₑ. The eyepiece magnifies it again.
  • Press “Final image at infinity”. The object moves a little; m drops to about 13.5.
  • Drag the object distance slider. If I₁ goes beyond Fₑ, the final image becomes real and useless to the eye.

Magnifying glass

A magnifying glass has a focal length of 5 cm. What is its magnification when the image is formed at the near point (25 cm)?

  1. 5
  2. 6
  3. 4
  4. 0.2
Answer

B. With the image at the near point, m = 1 + D/f = 1 + 25/5 = 6. (5 = D/f is the magnification with the image at infinity, for a relaxed eye.)

Compound microscope

A compound microscope has an objective of focal length 1.0 cm, an eyepiece of focal length 2.0 cm and a tube length of 20 cm. What is its magnifying power with the final image at infinity?

  1. 250
  2. 270
  3. 40
  4. 12.5
Answer

A. m = mₒ mₑ = (L/fₒ)(D/fₑ) = (20/1.0) × (25/2.0) = 20 × 12.5 = 250. (270 uses mₑ = 1 + D/fₑ, for the final image at the near point; 40 is L × fₑ/fₒ, not a magnification.)

Summary

  • Simple microscope: m = 1 + D/f (image at D); m = D/f (image at ∞)
  • Objective: mₒ = L/fₒ
  • Compound microscope: m = mₒ mₑ ≈ (L/fₒ)(D/fₑ)
  • Apparent size depends on the angle at the eye; the near point is D ≈ 25 cm.
  • A simple microscope is a convex lens with the object inside its focus: virtual, upright, magnified image.
  • In a compound microscope the objective makes a real, magnified image, which the eyepiece magnifies further.
  • Short focal lengths for both lenses and a long tube give high magnification.