AQA A Level Physics

Revision Notes

9.1.1 Lenses & Ray Diagrams for Telescopes

Lenses & Ray Diagrams for Telescopes

  • A lens is a piece of equipment that forms an image by refracting light
  • There are two types of lenses: 
    • A convex, or converging lens
    • A concave, or diverging lens
  • Note: in the Astrophysics module, only converging lenses will be discussed
  • In a converging lens, parallel rays of light are brought to a focus along the principal axis
    • This point is called the focal point
  • The distance from the centre of the lens to the focal point is called the focal length
    • This length depends on how curved, or how thick, the lens is
    • The more curved (thicker) the lens, the shorter the focal length
    • The shorter the focal length, the more powerful the lens

Lens Power

The focal length is shorter in a lens that is thicker and more curved. This makes for a more powerful lens

Real & Virtual Images

  • Images produced by lenses can be either real or virtual
Real image Virtual image
light converges towards a focal point light diverges away from a focal point
always inverted always upright
can be projected onto a screen cannot be projected onto a screen
intersection of two solid lines intersection of two dashed lines (or a dashed and a solid line)

Real Image, downloadable IGCSE & GCSE Physics revision notes

example: image from a projector onto a screen

Virtual Image

example: image in a mirror

Constructing Ray Diagrams

  • When constructing ray diagrams of refractors, it is generally assumed that the lenses used are very thin
    • This simplifies the situation by reducing the amount the incident rays of light refract
  • As a result, the three main rules for constructing ray diagrams are as follows:

1. Rays passing through the principal axis will pass through the optical centre of the lens undeviated

9-1-1-ray-diagram-construction-rule-1

2. Rays that are parallel to the principal axis will be refracted and pass through the focal point f

9-1-1-ray-diagram-construction-rule-2

3. Rays passing through the focal point f will emerge parallel to the principal axis

9-1-1-ray-diagram-construction-rule-3

Image Formation by a Converging Lens

  • Images formed by lenses can be described by their
    • Nature: Real or virtual
    • Orientation: Inverted or upright (compared to the object)
    • Size: Magnified (larger), diminished (smaller), or the same size (compared to the object)

Drawing ray diagrams of real images

  • For an object placed at a distance greater than 2 focal lengths...

pEJ3dJOx_9-1-1-lens-image-distance-greater-than-2f

  • The image that forms will have the following properties:
The image forms... between f and 2f
The nature of the image is... real
The orientation of the image is... inverted
The size of the image is... diminished

 

  • For an object placed at a distance equal to 2 focal lengths...

9-1-1-lens-image-distance-equal-to-2f

  • The image that forms will have the following properties:
The image forms... at 2f
The nature of the image is... real
The orientation of the image is... inverted
The size of the image is... the same

 

  • For an object placed at a distance between 1 and 2 focal lengths

9-1-1-lens-image-distance-greater-than-2f

  • The image that forms will have the following properties:
The image forms... beyond 2f
The nature of the image is... real
The orientation of the image is... inverted
The size of the image is... magnified

 

Drawing ray diagrams of virtual images

  • For an object placed at a distance less than the focal length (i.e. a magnifying glass):

9-1-1-lens-image-distance-less-than-f

  • The image that forms will have the following properties:
The image forms... at 2f (on the same side as the object)
The nature of the image is... virtual
The orientation of the image is... upright
The size of the image is... magnified

Worked example

Draw a ray diagram to show how a converging lens can be used to form a diminished image of a real object.

Label the object, image and principal foci of the lens on your diagram.

Answer:

Step 1: Start by drawing and labelling a principal axis and the lens as a line or a very thin ellipse

2N6Wi3i3_9-1-1-we-converging-lens-ray-diagram-step-1

Step 2: Mark and label the focal points on each side of the lens

hTVLIwle_9-1-1-we-converging-lens-ray-diagram-step-2

Step 3: Draw and label the object at a distance greater than the focal length on the left side of the lens

8KHMUsig_9-1-1-we-converging-lens-ray-diagram-step-3

  • Tip: the object should be placed a distance of at least 2F away from the lens

Step 4: Draw a ray through the optical centre of the lens 

TiD0zx2q_9-1-1-we-converging-lens-ray-diagram-step-4

Step 5: Draw a second ray from the object to the lens which is parallel to the principal axis

f2Ta3-ii_9-1-1-we-converging-lens-ray-diagram-step-5

Step 6: Draw the continuation of the ray passing through the focal point on the right side of the lens

aR17_81V_9-1-1-we-converging-lens-ray-diagram-step-6

Step 7: Draw and label the image at the point where the rays meet

1BTAMSH5_9-1-1-we-converging-lens-ray-diagram-step-7

Step 8: Check your final image and make sure everything is included to gain the marks

ltX4J6bI_9-1-1-we-converging-lens-ray-diagram-step-8

  • For a three-mark question, examiners will be looking for:
    • One ray drawn through the optical centre of the lens
    • A second ray drawn which produces a diminished (smaller) image (which must pass through a labelled focal point)
    • Both the object and the image must be drawn and labelled correctly

Exam Tip

When drawing ray diagrams, convex (converging) and concave (diverging) lenses can be simplified using the following symbols:

convex-concave-symbols-igcse-and-gcse-physics-revision-notes

The Lens Equation & Magnification

The Lens Equation

  • The lens equation relates the focal length of a lens to the distances between the lens and the image and the object

1 over f space equals space 1 over u space plus space 1 over v

  • Where:
    • = focal length (m)
    • v = distance of the image from lens (m)
    • u = distance of the object from lens (m)

Lens Equation for a Real Image

9-1-1-lens-equation-real-image

All values are positive for a real image

  • This equation only works for thin converging (or diverging) lenses
    • If the image is real, the value of v is positive 
    • If the image is virtual, the value of v is negative

Lens Equation for a Virtual Image

9-1-1-lens-equation-virtual-image

A virtual image forms on the same side as the object making the value of v negative

Magnification as a Ratio of Heights

  • Magnification, M, is the ratio of the image height to the object height

M space equals space h subscript i over h subscript o

  • Where:
    • M = magnification
    • h subscript i = image height (m)
    • h subscript o = object height (m)

Magnification as a Ratio of Distances

  • A diagram of an object and its real image will produce similar triangles
  • Therefore, magnification can be determined by comparing the distances from the lens to the object and the image

Magnification Ray Diagram for a Real Image

5-23-magnification-derivation-edexcel-al-physics-rn-1

Magnification for a real image can be derived using similar triangles

  • This also works for virtual images

Magnification Ray Diagram for a Virtual Image

5-23-magnification-derivation-2_edexcel-al-physics-rn

Magnification for a virtual image can also be derived using similar triangles

  • As magnification can be described using the ratio of the two opposite sides h subscript i and h subscript o, the same ratio must apply to the two adjacent sides v and u
  • Therefore, magnification can also be written as:

 M space equals space v over u

  • Where:
    • M = magnification
    • v = distance from lens to image (m)
    • u = distance from lens to object (m)
  • Since magnification is a ratio, it has no units

Worked example

A magnifying glass has a focal length of 15 cm. It is held 5 cm away from a component which is being examined. 

Determine the magnification of the image.

Answer:

Step 1: Write the known values

  • Focal length, f = 15 cm
  • Distance between object and lens, u = 5 cm

Step 2: Use the lens formula and rearrange to make v the subject

1 over f space equals space 1 over u space plus space 1 over v

v space equals space open parentheses 1 over f space minus space 1 over u close parentheses to the power of negative 1 end exponent

v space equals space open parentheses 1 over 15 space minus space 1 fifth close parentheses to the power of negative 1 end exponent space equals space minus 7.5 space cm

  • The negative sign indicates a virtual image is formed (expected for a magnifying glass) and can be ignored for the next step

Step 3: Use the magnification formula to find the magnification of the image

M space equals space v over u space equals space fraction numerator 7.5 over denominator space 5 end fraction space equals space 1.5

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