DSE Physics | Lenses and Optical Instruments Lesson-Prep Guide (Tutor Edition) | Thin Lens Equation, Magnification and Vision Correction

2026-11-03

Tutor lesson-prep guides · Secondary Physics

Students can recite 1/f = 1/u + 1/v and still fall apart at a minus sign. One f = 10 cm lens and three object distances fix the equation, the magnification and the virtual image in one lesson.

Objectives — Students can: (1) use 1/f = 1/u + 1/v to find the image distance and use the sign of v to decide real or virtual; (2) use m = -v/u for magnification and explain what the sign and the size each mean; (3) convert between power and focal length with D = 1/f and explain myopia versus hyperopia correction.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Reciprocals	How to do 1/10 – 1/30	Common denominator, then invert
Sign convention	Sign of v for real and virtual	Real v > 0, virtual v < 0
Unit conversion	25 cm in metres	Divide by 100

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Three rays	Parallel through focus, through optical centre	Draw the crossing point
8-20 min	Thin lens equation	u = 30, f = 10	v = 15 cm
20-30 min	Magnification	m = -v/u	m = -0.5, inverted, smaller
30-42 min	Object inside focus	u = 5, f = 10	v = -10, m = +2, virtual
42-52 min	Power and correction	D = 1/f	Myopia -4.0 D, f = -25 cm
52-60 min	Exit check	Three questions marked live	All three right

03 Teaching script (read it out)

Scene 1: One equation, three different answers

Take one convex lens with f = 10 cm and change only the object distance.
u = 30 cm: 1/v = 1/10 – 1/30 = 3/30 – 1/30 = 2/30, so v = 15 cm: a real image 15 cm behind the lens.
u = 15 cm: 1/v = 1/10 – 1/15 = 1/30, so v = 30 cm and m = -2: real, inverted, twice the size.
u = 5 cm, object inside the focus: 1/v = 1/10 – 1/5 = -1/10, so v = -10 cm. The minus sign places the image on the same side as the object, so it is virtual, and m = +2 means upright and magnified. That is the magnifying glass.
Same equation, same lens, different object distance, opposite kind of image.

Students must say: why v grows from 15 to 30 as u falls from 30 to 15.

Scene 2: The sign in the magnification carries information

m = -v/u is not a bare number: the sign is part of the answer.
u = 30, v = 15 gives m = -0.5: negative means inverted, 0.5 means half size, so the image is real, inverted and diminished.
u = 15, v = 30 gives m = -2: inverted, twice the size, real. This is where most students lose marks, because intuition equates magnified with upright.
u = 5, v = -10 gives m = +2: upright, magnified, virtual. Put the three side by side and the sign rule becomes a pattern rather than a memory test.

Students must say: what m = -2 means in full (inverted, twice the size, real, catchable on a screen).

Scene 3: Power links focal length to correction

Power D = 1/f with f in metres. A convex lens of focal length 25 cm has D = 1/0.25 = +4.0 D.
A myopic eye has a far point at 25 cm, so it needs a concave lens of f = -25 cm (D = -4.0 D) that images infinity at the far point. Hyperopia takes a convex lens instead, pulling a near object back onto the retina.
For instruments: microscope magnification is objective times eyepiece, 40x with 10x giving 400x. The objective forms a magnified real image and the eyepiece then acts as a magnifying glass on that image, so there are two stages and they must be described separately.

Students must say: why myopia takes a concave lens and hyperopia a convex one (where the image lands).

04 Question types and worked steps

Type	Example	Steps	Answer
Calculation	Image distance and nature	1/f = 1/u + 1/v, read the sign of v	15 cm, real
Calculation	Magnification	m = -v/u, sign gives orientation	-0.5
Calculation	Power and focal length	D = 1/f, convert cm to m first	-4.0 D, f = -25 cm
Explanation	Images in optical instruments	Objective real image, eyepiece as magnifier	Both stages described

Sanity check (teach this): Fix the units first: f and u in cm give v in cm.
Use a common denominator: 1/10 – 1/30 = 3/30 – 1/30 = 2/30, so v = 15 and not 20.
A negative v is virtual and on the same side as the object; never answer “15 cm behind the lens”.
Power uses metres: 25 cm is 0.25 m, so D = 4.0 and not 0.04.

05 Independent practice (3 questions)

Question	Answer	What to watch
f = 10, u = 30 image distance	15 cm, real	Subtract reciprocals, then invert
f = 10, u = 15 magnification	-2, inverted and magnified	m = -v/u
Power of a 25 cm focal length	+4.0 D	D = 1/0.25

06 Common mistakes and what to say

Mistake	Student writes	Say this
No common denominator	v = 20 cm	“1/10 minus 1/30 is 2/30, so v is 15.”
Calling a virtual image real	u = 5 answered as 10 cm behind the lens	“A negative v puts the image on the object side. It is virtual.”
Dropping the minus in m	m = 0.5	“The minus is what tells you the image is inverted.”
Power in centimetres	D = 1/25 = 0.04	“D needs f in metres, and 25 cm is 0.25 m.”

07 Homework (paper, section and question numbers)

Secondary physics chapter index — Lens and optics practice
All lesson-prep guides — Compare other subjects

08 Exit check (last 5 minutes)

Question	Answer	Decision
Image distance at u = 30	15 cm	All three right: move to instruments
Magnification at u = 15	-2	One wrong: reteach m = -v/u
Power of a 25 cm lens	+4.0 D	Two wrong: return to act three

09 Tutor FAQ

Must students learn the three special rays?

Yes: parallel through the focus, through the centre, and through the focus to parallel. Ray diagrams depend on them.

Can a screen catch a virtual image?

No. A virtual image is where backward extensions meet, so it is seen from the front of the lens.

Why is power defined with metres?

D is 1/f with f in metres by definition, so centimetres are out by a factor of 100.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.