Refracting Telescopes (AQA A Level Physics): Revision Note

Draw a ray diagram for an astronomical refracting telescope in normal adjustment.

Your diagram should show the paths of three non-axial rays passing through both lenses.

Label the principal foci of the two lenses.

[3]

Answer:

Step 1: Start by drawing and labelling the two lenses

• If the question does not include a principal axis, make sure to draw this first

• Draw the lenses as lines with arrows or very thin ellipses

Step 2: Mark and label the common principal foci

• The objective focal length must be longer than the eyepiece focal length $\left(f_o>f_e\right)$ [1 mark]

• You could also mark a single point and label it F 

Step 3: Draw an off-axis ray through the centre of the objective to the eyepiece

• The rays must be off-axis (non-axial), meaning drawn at an angle to the principal axis

Step 4: Draw an arrow to show the intermediate image from the common principal foci to this ray

• You don't have to do this step, but it will help the examiner to clearly see your working

Step 5: Draw a construction line from the end of the intermediate image through the centre of the eyepiece

• You don't have to do this either, but it will help the examiner to clearly see your working

Step 6: Draw two rays to the eyepiece, crossing at where the focal lengths meet

[1 mark]

Step 7: Draw the continuation of the three rays from the eyepiece, parallel to the construction line

[1 mark]

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

• For a three-mark question, examiners will be looking for:

  • Both focal points are marked and labelled at the same point on the principal axis with $f_o>f_e$

  • Three off-axis rays drawn through the objective lens

  • Three rays drawn through the eyepiece lens parallel to a construction line

It is important that you get lots of practice drawing this diagram. It's a common exam question, and examiners can be extremely meticulous when it comes to marking them.

Make sure to avoid these common problems when drawing the diagram:

• Drawing axial rays (parallel to the principal axis) rather than non-axial rays (at an angle to the principal axis)

• Bending the central ray at the objective lens

• Bending the rays at the intermediate image (i.e. at $f_o$, $f_e$)

• Not drawing the rays parallel to each other, or the construction line, at the eyepiece lens

• Labelling the principal foci where the rays cross, rather than on the principal axis

A student is looking for suitable lenses to use in a simple refracting astronomical telescope. 

The student sets out to measure the focal length of a converging lens by placing an object and screen at a fixed distance of 300 cm apart.

A sharp image is observed on the screen when the lens is placed between the object and the screen at a distance of 207 cm from the object.

(a) Calculate the focal length of the lens. [3]

(b) State whether the lens formed the eyepiece or objective, giving reasons for your answer. [2]

Answer:

Part (a)

Step 1: Sketch a quick diagram to visualise the scenario

Step 2: Write down the lens equation

lens equation:  $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

• Where $u=207\mathrm{cm}$ and $v=93\mathrm{cm}$ [1 mark]

Step 3: Calculate the focal length of the lens

$\frac{1}{f}=\frac{1}{207}+\frac{1}{93}$ [1 mark]

$f={\left(\frac{1}{207}+\frac{1}{93}\right)}^{-1}=64\mathrm{cm}$ [1 mark]

Part (b)

You could either argue...

• The eyepiece forms a virtual, inverted, magnified image

• The objective lens forms a real, inverted, diminished image of a distant object very near the focal point of the eyepiece [1 mark]

• The image formed by the lens in (a) is real, inverted, and diminished

• Therefore, the lens must be an objective lens [1 mark]

Or, you could argue...

• In a refracting telescope, angular magnification is given by:  $M=\frac{f_o}{f_e}$

• Hence, the eyepiece must have a shorter focal length and the objective must have a longer focal length:  $f_o>f_e$

• For the best possible magnification, the image due to the objective lens, which acts as the object for the eyepiece lens, must be located at the focal point of the eyepiece [1 mark]

• Hence, the two lenses must be separated by a distance $f_o+f_e$ (i.e. the telescope length)

• Therefore, the lens must be an objective lens [1 mark]

An astronomical refracting telescope which is 1.23 m long and has an angular magnification of 200 is used to observe Neptune at its closest approach to Earth.

• distance from Earth to Neptune at closest approach = 4.3 × 10⁹ km

• diameter of Neptune = 4.9 × 10⁴ km

(a) Calculate the focal lengths of the objective lens and eyepiece lens. [3]

(b) Calculate the angle subtended by the image of Neptune when viewed through this telescope. [2]

Answer:

Part (a)

Step 1: List the relevant equations and known quantities

Telescope length:  $f_o+f_e=1.23\mathrm{m}$

Angular magnification:  $M=\frac{f_o}{f_e}=200$

Step 2: Write an expression for one of the focal lengths in terms of the other

$f_o=200f_e$ [1 mark]

$200f_e+f_e=1.23\mathrm{m}$

$201f_e=1.23\mathrm{m}$

Step 3: Solve to determine the focal lengths of the lens

• Focal length of the eyepiece:  $f_e=\frac{1.23}{201}=6.1\times {10}^{-3}\mathrm{m}\approx 0.01\mathrm{m}$ [1 mark]

• Focal length of the objective:  $f_o=200\times \left(6.1\times {10}^{-3}\right)=1.22\mathrm{m}$ [1 mark]

Part (b)

Step 1: Recall the equations for angular size and angular magnification

Angular size: $\theta =\frac{h}{d}$

• Where diameter, $h=4.9\times {10}^4\mathrm{km}$ and distance, $d=4.3\times {10}^9\mathrm{km}$

Angular magnification:  $M=\frac{\beta }{\alpha }$

• Where $\alpha$ is the angle subtended by Neptune at unaided eye and $\beta$ is the angle subtended by the image of Neptune at eye

Step 2: Determine the angle subtended by Neptune at unaided eye

$\theta =\frac{4.9\times {10}^4}{4.3\times {10}^9}=1.14\times {10}^{-5}\mathrm{rad}$ [1 mark]

• The angle subtended by Neptune with the unaided eye: $\alpha =1.14\times {10}^{-5}\mathrm{rad}$

Step 3: Determine the angle subtended by the image of Neptune at the eyepiece

$M=\frac{\beta }{\alpha }\Rightarrow \beta =\alpha M$

$\beta =\left(1.14\times {10}^{-5}\right)\times 200=2.28\times {10}^{-3}\mathrm{rad}$ [1 mark]

• Angle subtended by the image of Neptune at eyepiece: $\beta =2.28\times {10}^{-3}\mathrm{rad}$