Lenses and Optical Instruments

Thin lens equation 1/f = 1/u + 1/v; magnification; real and virtual images

# Lenses and Optical Instruments — A-Level Physics

At A-Level, lens optics requires quantitative analysis using the thin lens equation and real/virtual image conventions.


1. The Thin Lens Equation

1f=1u+1v\boxed{\frac{1}{f} = \frac{1}{u} + \frac{1}{v}}

Where:

  • ff = focal length (m)
  • uu = object distance (m)
  • vv = image distance (m)

Sign Convention (Real-is-Positive)

  • Real objects: uu is positive (light enters from left)
  • Real images: vv is positive (form on opposite side to object)
  • Virtual images: vv is negative (form on same side as object)
  • Converging lens: ff is positive
  • Diverging lens: ff is negative

2. Magnification

m=vu=image heightobject height\boxed{m = \frac{v}{u} = \frac{\text{image height}}{\text{object height}}}

  • m>1|m| > 1: magnified
  • m<1|m| < 1: diminished
  • m>0m > 0: upright (virtual)
  • m<0m < 0: inverted (real)

3. Power of a Lens

P=1fP = \frac{1}{f}

Units: dioptres (D). Converging: positive power. Diverging: negative power.

For lenses in contact: Ptotal=P1+P2P_{\text{total}} = P_1 + P_2

Worked Example: Example 1

Problem

An object is 30 cm from a converging lens of focal length 20 cm. Find the image distance and magnification.

1/v=1/f1/u=1/201/30=3/602/60=1/601/v = 1/f - 1/u = 1/20 - 1/30 = 3/60 - 2/60 = 1/60 v=60v = 60 cm (positive → real image) m=v/u=60/30=2m = v/u = 60/30 = 2 (magnified, but using sign convention properly with real-is-positive: m=v/u=60/30=2m = -v/u = -60/30 = -2, meaning inverted and magnified)

Solution

Worked Example: Example 2

Problem

An object is 10 cm from a converging lens of focal length 15 cm.

1/v=1/151/10=2/303/30=1/301/v = 1/15 - 1/10 = 2/30 - 3/30 = -1/30 v=30v = -30 cm (negative → virtual image, same side as object) m=v/u=30/10=3m = v/u = 30/10 = 3 (upright, magnified — this is a magnifying glass)

Solution

Worked Example: Diverging Lens

Problem

An object is 20 cm from a diverging lens (f=15f = -15 cm).

1/v=1/(15)1/20=4/603/60=7/601/v = 1/(-15) - 1/20 = -4/60 - 3/60 = -7/60 v=8.57v = -8.57 cm (virtual, upright, diminished) m=8.57/20=0.43m = 8.57/20 = 0.43

Solution

5. Two-Lens Systems

The image from the first lens becomes the object for the second lens.

6. Practice Questions

    1. A converging lens (f=12f = 12 cm) forms an image of an object 18 cm away. Find image position and magnification. (3 marks)
    1. A diverging lens (f=20f = -20 cm) has an object 30 cm away. Find the image position. (2 marks)
    1. Two thin lenses of power +5 D and −2 D are in contact. Find the combined focal length. (2 marks)

    Answers

    1. 1/v=1/121/18=3/362/36=1/361/v = 1/12 - 1/18 = 3/36 - 2/36 = 1/36. v=36v = 36 cm. m=36/18=2m = 36/18 = 2 (real, inverted, magnified).
    1. 1/v=1/201/30=3/602/60=5/601/v = -1/20 - 1/30 = -3/60 - 2/60 = -5/60. v=12v = -12 cm (virtual).

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Summary

  • Thin lens equation: 1/f=1/u+1/v1/f = 1/u + 1/v
  • Sign convention: real positive, virtual negative
  • m=v/um = v/u; power P=1/fP = 1/f (dioptres)
  • Converging: f>0f > 0; Diverging: f<0f < 0

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