Investigating rays and optical images
Key idea: Trace rays and measure angles from the normal to test reflection and refraction. Locate images by sharp focus or no parallax, and use a distant object to estimate focal length.
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Guided lessons · coming soon
Guided lesson
Guided steps
- Locate the boundary and normal before measuring reflection or refraction.
- Convert the surface angle to the normal angle before applying reflection.
- Follow the sine ratio for air and glass.
- Read how to locate real and virtual images.
- Correct the distance, then check image size and type.
- Read how to find the critical angle and estimate focal length.
- Plan the semicircular-block method and state both TIR conditions.
- Compare the critical-angle cases and the reversed media.
- Explain the lamp-image mistake and plan a focal-length measurement.
- Recover the normal, image type and model conditions before tracing rays.
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ContinueA ray drawn neatly can still give a wrong angle if its reference line is wrong. In optical investigations, first locate the boundary, point of incidence and normal. For images, first decide whether rays actually meet or only appear to come from a position.
Follow instructions for ray boxes and other classroom sources. Never look into a laser beam or point it at someone. Never use the Sun for a lens experiment. Handle glass blocks and lenses as instructed, keeping clear of sharp edges.
Trace rays before measuring angles
Mark the incident and reflected paths using well-separated points. Draw the normal perpendicular to the mirror at the incidence point. Compare incidence and reflection angles, both measured from that normal, over several trials. An angle 30° from the surface is 60° from the normal.
For refraction, draw the block outline, mark entry and exit points, and join the internal path after removing the block. Use the normal at the same entry boundary to measure incidence in air and refraction in glass. Keep the block and ray alignment consistent and compare several angle pairs.
The law of reflection
Method
Direct a narrow ray at a plane mirror, mark the incident and reflected rays, draw the normal at the point of incidence, and measure angles from the normal. Repeat for several incidence angles and compare i with r.
Why it works
The law is tested with paired angles measured from the normal, not the mirror. A narrow ray, fixed mirror line, accurate point marking and repeated angles reduce tracing uncertainty.
A mistake to avoid: Angles of incidence and reflection are measured from the mirror surface.
Check your understanding
From which line must you measure both the angle of incidence and the angle of reflection?
Show answer
Measure both angles from the normal: the line perpendicular to the mirror at the point where the ray strikes. Repeat at several angles; agreement within tracing and angle-reading uncertainty supports i = r.
The refraction of light through glass blocks
Method
Trace a narrow ray through a glass block, mark entry and exit points, replace the block outline, and draw normals. Measure incidence and refraction angles from the normal for several trials; if required, compare sin i with sin r.
Why it works
Accurate boundary points and normals are essential because refractive angles are defined at the interface. Multiple angle pairs can test a constant sine ratio for the same two media.
A mistake to avoid: Refraction angles are measured between the ray and the glass surface.
Check your understanding
Which two angles would you record for each ray, and why is one pair of readings not enough?
Show answer
Record incidence in air and refraction in glass, both from the normal at the same boundary. Repeat at several incidence angles to see whether the sine ratio stays constant for the same media.
Check your understanding 1 Measure angles from the normal
A ray strikes a plane mirror at 20° to its surface. Find the incidence angle and the reflection angle, both using the correct reference line. Choose that reference line.
Show answer
The normal is 90° to the mirror surface, so i = 90° − 20° = 70°. The law of reflection gives r = 70°, also from the normal. Measuring r from the surface would instead give 20° and would use the wrong definition.
Worked example 1
Test a sine ratio
Problem
A ray enters glass from air. The angle to the normal in air is 40° and the angle to the normal in glass is 25°. Find sin i/sin r and explain how to test whether it is constant.
Show full solution
Use the defined angles
Method
Identify i in air and r in glass, both from the normal.
Reason
Measuring from the surface would substitute complementary angles into the wrong ratio.
Working
Answer
i = 40°, r = 25°. Use degree mode.Calculate and test
Method
Divide the two sines, then repeat with several different incidence angles.
Reason
One pair gives an estimate; several pairs test the relationship for the same media and light.
Working
Answer
sin 40°/sin 25° = 0.6428/0.4226 ≈ 1.52. Look for agreement within tracing and angle-reading uncertainty.
Quick check
A ratio greater than one is consistent with bending towards the normal on entering glass from air. It does not justify applying the same ratio to an air-to-water boundary.
Locate a real or virtual image
A plane-mirror image is virtual: reflected rays do not actually meet behind the mirror. Compare the image with a marker and move your eye sideways. Adjust the marker until their positions show no relative movement, which is the no-parallax condition. Measure from the mirror surface and compare object and image position, size and orientation.
A thin converging lens can make a real image on a screen when the object is beyond the focal length. Align object, optical centre and screen, then adjust for sharp focus. Measure distances from the lens’s optical centre and record image height and orientation as well as position. Inside the focal length, the image is virtual and cannot be focused on that screen.
Determination of the position and characteristics of an optical image formed by a plane mirror or a thin converging lens
Method
For a plane mirror, locate the virtual image by eliminating parallax between its image and a marker. For a thin converging lens, place the object beyond its focal length, focus the real image sharply on a screen, measure object and image distances from the optical centre, and record orientation and size at controlled object positions. An object inside the focal length forms a virtual image, so the screen method does not apply.
Why it works
A plane-mirror image cannot be caught on a screen, so a no-parallax comparison locates it. For a converging lens, place the object beyond the focal length to form a real screen image. Focus sharply and measure object and image distances from the lens's optical centre. An object inside the focal length gives a virtual image that cannot be focused on a screen.
A mistake to avoid: Every optical image can be located by placing a screen where the image appears to be.
Check your understanding
Why is a screen useful for a converging-lens image but not for a plane-mirror image?
Show answer
A real converging-lens image is formed where rays actually converge, so a screen can receive it. A plane-mirror image is virtual: the reflected rays only appear to come from behind the mirror. Locate it using a marker and no-parallax comparison, rather than a screen.
Guided practice 2
Find a real image and its characteristics
Problem
A lens has focal length 10 cm and thickness 1.2 cm. Its optical centre is halfway through that thickness. A 24 mm tall object is 50 cm from the optical centre. A sharp, 6.0 mm tall image forms on a screen 11.9 cm from the lens’s back surface. Find the centre-to-back-surface distance and correct a learner’s image distance of 11.9 cm. Give both distances to one decimal place. Give the ratio of image height to object height and choose its orientation. Could a sharp screen image still be found if the object moved to 6.0 cm from the centre?
Try before revealing the solution
Show the steps Optional working
Hints
Hint 1
Find the decisive relationship
Measure from the optical centre: add half the thickness to the surface-to-screen gap. Compare the two heights using the same unit.
Show solution step by step
Locate the image
Method
Measure from the optical centre rather than the back surface.
Reason
The optical centre is half the lens thickness farther from the screen than the back surface.
Working
Answer
Half the thickness is 1.2/2 = 0.60 cm. The image distance is 11.9 + 0.60 = 12.5 cm, between f = 10 cm and 2f = 20 cm.
Record characteristics
Method
Compare the focused image with the object.Reason
For an object beyond 2f, the image is between f and 2f, inverted and smaller.
Working
Answer
The image is real, inverted and diminished: its height is 6.0/24 = 0.25 times the object’s height.
Identify the boundary
Method
Compare the new 6.0 cm object distance with f = 10 cm.
Reason
With the object inside f, the outgoing rays diverge; no screen on the far side produces a sharp image.
Working
Answer
At 6.0 cm the object is inside f. No screen adjustment can produce a sharp real image; the image is virtual, upright and magnified on the object side. Never use the Sun as the object.
Quick check
A plane-mirror image is also virtual. A marker can locate it by eliminating parallax; placing a screen at its apparent position cannot capture it.
Find the critical angle and estimate focal length
For a glass-to-air boundary, increase incidence until the refracted ray runs along the surface. This is the critical angle: the refracted angle is 90°. Total internal reflection occurs for incidence greater than this angle, when light travels from higher to lower refractive index. A semicircular block allows entry along a radius, avoiding bending at the curved entry face; the flat face is then the tested boundary.
For a focal-length estimate, focus a sufficiently distant illuminated object sharply on a screen. Its rays arrive nearly parallel, so the optical-centre-to-screen distance is approximately f. A sharp image of a nearby object does not generally give f directly. Repeat and evaluate the focus and distance readings.
Try it yourself 3
Use a semicircular block to find the critical angle
Problem
Describe how a semicircular glass block can be used to find the glass-to-air critical angle. State both conditions for total internal reflection and how to recognise the boundary angle.
Try before revealing the solution
Write the complete method on paper or here, then compare the steps and conditions with the model. You don’t need to type up an answer written on paper.
Hints
Hint 1
Choose the relevant evidence
The entry along a radius and the flat exit face have different jobs. Identify the normal and angle at the flat face.
Show solution step by step
Enter along a radius
Method
Send the narrow ray through the curved face towards the centre.
Reason
At the curved entry face the ray follows a radius, so incidence is normal and it is not bent there.
Working
Answer
Draw the normal to the flat face at the centre; measure the internal incidence angle from that normal.
Approach the boundary
Method
Increase incidence at the glass-to-air flat face until the refracted ray runs along the surface.
Reason
At the critical angle the angle of refraction is 90°. Beyond it, no transmitted ray emerges into the air.
Working
Answer
The critical angle is that internal angle. Total internal reflection requires travel from higher to lower refractive index and incidence greater than the critical angle.
Check rather than guess
Method
Take several readings close to the transition and trace the ray carefully.
Reason
A faint ray becoming hard to see is not a reliable criterion on its own.
Working
Answer
Use the surface-running ray and the internal normal to locate the boundary. For glass with n ≈ 1.52, an estimate near 41° is plausible.
Quick check
Compare the radial entry, the normal and boundary ray at the flat face, both TIR conditions, and repeated tracing near the transition. At the critical angle the ray is refracted along the boundary; total internal reflection occurs for incidence greater than that angle.
The principle of total internal reflection
Method
Send a ray through the curved face of a semicircular block so it reaches the flat face from the denser medium. Increase the incidence angle, observe refraction and reflection, and identify the critical angle when the refracted ray travels along the boundary.
Why it works
The curved entry face minimises unwanted refraction when the ray is aimed through the centre. Total internal reflection occurs only from higher to lower refractive index and above the critical angle.
A mistake to avoid: Total internal reflection occurs whenever light meets a boundary at a large angle from either side.
Check your understanding
State the two conditions needed for total internal reflection before describing how to find the critical angle.
Show answer
Light must travel from a higher to a lower refractive index, with incidence greater than the critical angle. Enter a semicircular block along a radius so the curved entry causes no bending. Find the angle at the flat face where the refracted ray lies along that face; that is the critical angle.
Check your understanding 2 At and above the critical angle
A ray travels from glass to air. Its incidence is 42° from the normal, and the critical angle is 41°. Choose what happens at 42° and at exactly 41°. Then choose why the same rule cannot simply be used for an air-to-glass ray.
Show answer
At 42°, incidence exceeds the critical angle on a higher-to-lower-index boundary, so total internal reflection occurs. At exactly 41°, the refracted ray travels along the boundary. From air to glass, light travels towards the higher refractive index, so the required index condition for total internal reflection is absent.
The focal length of lenses
Method
Form a sharp image of a distant object on a screen and measure from the lens's optical centre to the screen for an estimate of focal length. Repeat and average; for finite object distances, measure u and v and use the lens relation if required.
Why it works
Light from a distant object reaches the lens nearly parallel, so its sharp real image forms near the focal plane. Sharpness, distance reference and repeated readings dominate the uncertainty.
A mistake to avoid: Focal length is the distance from the object to the lens for any object position.
Check your understanding
Why does a distant object let the lens-to-screen distance estimate the focal length?
Show answer
Rays arriving from each point of a sufficiently distant object are nearly parallel at the lens, so they form an image near the focal plane. Focus that image sharply and measure from the optical centre to the screen. A nearby object’s image distance is not generally f. Never use the Sun.
Spot the mistake 4
Does every sharp image give the focal length?
Problem
A student forms a sharp image of a nearby lamp 30 cm beyond a lens and reports f = 30 cm. Explain the error and a method that gives a useful estimate of f.
Try before revealing the solution
Write the complete method on paper or here, then compare the steps and conditions with the model. You don’t need to type up an answer written on paper.
Show solution step by step
Image distance and focal length
Method
A finite-distance object generally produces a real image beyond the focal plane.
Reason
The lens-to-screen distance equals f only approximately when incoming rays are nearly parallel.
Working
Answer
A sharp image locates the image, but does not by itself show the object was sufficiently distant.
Choose a distant target
Method
Use a distant illuminated object, keep alignment fixed and move the screen for the sharpest image.
Reason
Rays from each point of a sufficiently distant object are approximately parallel at the lens.
Working
Answer
Measure the optical-centre-to-screen distance as an approximate f. Repeat and compare; never point the lens at the Sun.
Common mistake
Compare the choice of distant target, sharp focus, optical-centre distance and repeated readings. If a question supplies a finite-object lens equation, use its measured object and image distances. Do not silently treat a nearby object as infinitely distant.
Common mistakes
- Measuring angles from the surface instead of from the normal.
- Drawing conclusions about refraction from a single pair of angles.
- Trying to catch a plane-mirror image on a screen. It is virtual.
- Taking any sharp image distance as the focal length. Only an image of a distant object forms at about the focal length.
- Forgetting that total internal reflection needs light travelling towards the less optically dense medium.
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Syllabus and review details
- SEC G3 Physics 2027 · 2027
Content Structure, PDF page 9; Subject Content, PDF pages 10–28
Last reviewed: