This MCQ module is based on: Mirror Formula, Magnification and Numerical Problems
TOPIC 33 OF 50
Mirror Formula, Magnification and Numerical Problems
🎓 Class 10
Science
CBSE
Theory
Ch 9 — Light – Reflection and Refraction
⏱ ~21 min
🌐 Language: [gtranslate]
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Part 2 — Putting Numbers on the Ray Diagram
In Part 1 we drew ray diagrams to see where an image forms. Now we need equations to calculate the exact image distance and image size. Two expressions do the whole job — the mirror formula and the magnification formula — provided we apply the New Cartesian sign convention carefully.
9.6 The Mirror Formula
For a spherical mirror of focal length \(f\), if an object is placed at distance \(u\) from the pole along the principal axis and its image forms at distance \(v\), the three distances are related by:
Mirror Formula
\[ \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \]
where all distances are measured from the pole with their appropriate signs (New Cartesian convention).
Sign-convention reminders: Object is always on the left → \(u\) is negative. For a concave mirror \(f\) is negative; for a convex mirror \(f\) is positive. A real image (in front of mirror) gives negative \(v\); a virtual image (behind mirror) gives positive \(v\).
9.7 Linear Magnification
Magnification (\(m\)) tells us how big the image is compared with the object:
\[ m = \dfrac{\text{height of image}}{\text{height of object}} = \dfrac{h'}{h} = -\dfrac{v}{u} \]
- If \(|m|>1\) → image is enlarged; if \(|m|<1\) → image is diminished; if \(|m|=1\) → same size.
- A negative value of \(m\) means the image is real and inverted.
- A positive value of \(m\) means the image is virtual and erect.
9.8 Worked Numerical Examples
Example 1 — Concave mirror, object beyond C
Given: A concave mirror of focal length 15 cm. An object is placed 20 cm in front of the mirror. Find the image distance and describe the image.
Step 1 — Sign convention: \(f = -15\) cm (concave), \(u = -20\) cm.
Step 2 — Mirror formula:
\[ \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\quad\Rightarrow\quad \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{-15} - \dfrac{1}{-20} \] \[ \dfrac{1}{v} = -\dfrac{1}{15} + \dfrac{1}{20} = \dfrac{-4 + 3}{60} = -\dfrac{1}{60} \]\(\therefore v = -60\) cm.
Step 3 — Magnification: \(m = -v/u = -(-60)/(-20) = -3\).
Image: real (v negative), inverted (m negative), 3 times enlarged, formed 60 cm in front of the mirror (between C and beyond).
Example 2 — Concave mirror, object between P and F
Given: A concave mirror has \(f = 20\) cm. An object 4 cm tall is placed 10 cm in front of it. Find \(v\), \(m\) and \(h'\).
\(f = -20\) cm, \(u = -10\) cm, \(h = 4\) cm.
\[ \dfrac{1}{v} = \dfrac{1}{-20} - \dfrac{1}{-10} = -\dfrac{1}{20} + \dfrac{1}{10} = \dfrac{-1+2}{20} = \dfrac{1}{20} \]\(v = +20\) cm (virtual image behind mirror).
\(m = -v/u = -(20)/(-10) = +2\) → image is erect and enlarged twice.
\(h' = m \cdot h = 2 \times 4 = 8\) cm (erect).
This is the shaving-mirror case.
Example 3 — Convex mirror
Given: A convex mirror has radius of curvature 30 cm. An object is placed 20 cm in front of it. Find the image distance and magnification.
For a convex mirror \(f\) is positive: \(f = R/2 = +15\) cm. \(u = -20\) cm.
\[ \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{15} - \dfrac{1}{-20} = \dfrac{1}{15} + \dfrac{1}{20} = \dfrac{4+3}{60} = \dfrac{7}{60} \]\(v = 60/7 \approx +8.57\) cm (virtual, behind the mirror).
\(m = -v/u = -(8.57)/(-20) = +0.43\).
Image is virtual, erect and diminished to about 43% of the object size — typical behaviour of a convex mirror.
Example 4 — Find the focal length
Given: An object is placed 25 cm in front of a concave mirror and its real image is formed at 75 cm on the same side. Find \(f\) and \(R\) of the mirror.
\(u = -25\) cm, \(v = -75\) cm (real image is in front → negative).
\[ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{-75} + \dfrac{1}{-25} = -\dfrac{1}{75} - \dfrac{3}{75} = -\dfrac{4}{75} \]\(f = -75/4 = -18.75\) cm. Radius \(R = 2f = -37.5\) cm.
So the mirror is concave with focal length 18.75 cm and radius of curvature 37.5 cm.
Example 5 — Finding where to place the object
Given: A concave mirror of focal length 18 cm must form a real image three times the size of the object. Where should the object be placed?
Real + inverted → \(m = -3\). So \(-v/u = -3 \Rightarrow v = 3u\).
Also \(f = -18\) cm. Using the mirror formula:
\[ \dfrac{1}{3u} + \dfrac{1}{u} = \dfrac{1}{-18}\quad\Rightarrow\quad \dfrac{1 + 3}{3u} = -\dfrac{1}{18}\quad\Rightarrow\quad \dfrac{4}{3u} = -\dfrac{1}{18} \]\(3u = -72\) → \(u = -24\) cm. So the object must be placed 24 cm in front of the mirror. Image distance \(v = 3u = -72\) cm (real).
Example 6 — Image height of a convex mirror
Given: A 5 cm tall candle is placed 40 cm in front of a convex mirror whose focal length is 20 cm. Find the position, nature and size of the image.
\(f = +20\) cm, \(u = -40\) cm, \(h = 5\) cm.
\[ \dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{-40} = \dfrac{1}{20} + \dfrac{1}{40} = \dfrac{2+1}{40} = \dfrac{3}{40} \]\(v = 40/3 \approx 13.33\) cm (virtual, positive).
\(m = -v/u = -(13.33)/(-40) = +0.333\).
\(h' = m\cdot h = 0.333 \times 5 = 1.67\) cm.
Image is virtual, erect, 1.67 cm tall, 13.33 cm behind the mirror.
Example 7 — When v and u are swapped
Given: A concave mirror of focal length 10 cm forms an image of a 2 cm tall object placed 15 cm in front of it. Find image position, magnification and image height.
\(f = -10\) cm, \(u = -15\) cm.
\[ \dfrac{1}{v} = \dfrac{1}{-10} - \dfrac{1}{-15} = -\dfrac{1}{10} + \dfrac{1}{15} = \dfrac{-3+2}{30} = -\dfrac{1}{30} \]\(v = -30\) cm (real, inverted, in front of mirror).
\(m = -v/u = -(-30)/(-15) = -2\).
\(h' = -2 \times 2 = -4\) cm (negative → inverted).
Object between C and F, image beyond C, enlarged, real and inverted — exactly Case 4 from Part 1.
Example 8 — Magnification from two measurements
Given: A pencil of length 4 cm placed along the axis produces an image 12 cm long. Find the magnification and state the type of mirror.
\(m = h'/h = 12/4 = 3\) (in magnitude).
If the image is real and inverted (as in a projector), \(m = -3\); the mirror must be concave and the object must lie between C and F. If the image is erect (as in a shaving mirror), \(m = +3\); again a concave mirror is needed with the object between P and F. A convex mirror can never give \(|m|>1\), so this cannot be a convex mirror.
Activity 9.2 — Measure the Focal Length of a Concave Mirror Using a CandleL3 Apply
Aim: To determine the focal length of a given concave mirror by obtaining a sharp image of a lighted candle on a screen and using the mirror formula.
Materials: Concave mirror with stand, candle, white paper screen, metre-scale, a darkened room.
Procedure:
- Place the candle on a table with its flame slightly above the level of the mirror's pole.
- Move the screen back and forth along the principal axis until a sharp, inverted image of the flame appears on it.
- Measure object distance \(u\) (candle to mirror) and image distance \(v\) (mirror to screen), both with a minus sign.
- Repeat for three different object distances (25 cm, 30 cm, 40 cm) and calculate \(1/f = 1/v + 1/u\) each time. Take the mean.
Predict: Will your three readings of \(f\) be exactly equal? What kinds of error might make them differ?
In principle the three values of \(f\) must match. In practice small disagreements arise from (i) difficulty in catching the sharpest image, (ii) mirror-aperture not being truly small and (iii) the candle flame being an extended source rather than a point. Averaging many readings reduces random error and gives a reliable focal length. This is the standard school-lab method of measuring \(f\) of a concave mirror.
Interactive — Mirror Formula Calculator
Enter any two of \(u\) (cm), \(f\) (cm) and height \(h\) (cm) with correct sign. The tool finds \(v\), \(m\) and \(h'\).
u (cm):
f (cm):
h (cm):
Values shown: concave mirror example.
Competency-Based Questions
In her physics practical, Meera places an object 30 cm in front of a concave mirror of focal length 20 cm. She finds that the image on her screen is inverted and bigger than the object. A friend insists the image should be smaller because the object is "far from the focus". Meera checks her calculation.
Q1. Using the mirror formula, find \(v\) in Meera's set-up. L3 Apply
\(u=-30\) cm, \(f=-20\) cm. \(1/v = -1/20 - (-1/30) = -1/20 + 1/30 = (-3+2)/60 = -1/60\). So \(v = -60\) cm — a real image 60 cm in front.
Q2. Calculate the magnification. Who is right — Meera or her friend? L4 Analyse
\(m = -v/u = -(-60)/(-30) = -2\). Magnitude 2 ⇒ image is twice as big as the object, inverted. Meera is right. The object lies between C (40 cm) and F (20 cm), so Case 4 applies.
Q3. (MCQ) The magnification of a plane mirror is always L1 Remember
(d) +1. The image in a plane mirror is virtual, erect and of the same size as the object.
Q4. An object is placed 15 cm in front of a convex mirror of radius 30 cm. Is the image virtual or real? Find \(v\) and \(m\). L3 Apply
\(f = R/2 = +15\) cm, \(u=-15\) cm. \(1/v = 1/15 - 1/(-15) = 1/15 + 1/15 = 2/15\). \(v=+7.5\) cm (virtual). \(m = -v/u = -7.5/(-15) = +0.5\). Image is virtual, erect, half the size.
Q5. (Short answer) Why does a convex mirror always give \(|m|<1\), no matter where the object is placed? L4 Analyse
For a convex mirror \(f>0\) and \(u<0\). The mirror formula forces \(0
Assertion–Reason Questions
Options: (A) Both A & R true, R correctly explains A. (B) Both A & R true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
Assertion (A): For a real image formed by a concave mirror, the magnification is negative.
Reason (R): For a real image, both \(u\) and \(v\) are negative, so \(m = -v/u\) comes out negative.
(A) — Both true; R correctly explains A. Negative \(m\) indicates inverted, real image.
Assertion (A): A virtual image formed by a concave mirror is always enlarged.
Reason (R): For an object between P and F of a concave mirror, \(|v|>|u|\), so \(|m|>1\).
(A) — Both true; R correctly explains A.
Assertion (A): The focal length of a convex mirror is taken as positive.
Reason (R): The principal focus of a convex mirror lies behind the mirror, on the same side as the outgoing (positive) direction of light after reflection.
(A) — Both true; R correctly explains A. In the New Cartesian convention distances to the right of the pole are positive.
Frequently Asked Questions — Mirror Formula & Magnification
What is mirror formula & magnification in Class 10 Science (CBSE board)?
Mirror Formula & Magnification is a key topic in NCERT Class 10 Science Chapter 9 — Light - Reflection and Refraction. It explains cartesian sign convention, mirror formula 1/v + 1/u = 1/f and magnification for spherical mirrors with numerical problems. Core ideas covered include mirror formula, magnification, sign convention, focal length. Mastering this subtopic is essential for scoring well in the CBSE Class 10 Science board exam because board papers repeatedly test these concepts through MCQs, short answers and long-answer questions. This part gives a complete, exam-ready explanation with activities, diagrams and competency-based practice aligned to NCERT.
Why is mirror formula important in NCERT Class 10 Science?
Mirror formula is important in NCERT Class 10 Science because it forms the foundation for understanding mirror formula & magnification in Chapter 9 — Light - Reflection and Refraction. Without a clear idea of mirror formula, students cannot answer higher-order CBSE board questions involving magnification, sign convention, focal length. Board papers regularly include 2-mark and 3-mark questions on this concept, and competency-based questions often link mirror formula to real-life situations. Building clarity here pays off directly in board marks.
How is mirror formula & magnification tested in the Class 10 Science CBSE board exam?
The CBSE Class 10 Science board exam tests mirror formula & magnification through a mix of 1-mark MCQs, 2-mark short answers, 3-mark explanations with examples, 5-mark descriptive questions (often with diagrams or balanced equations) and 4-mark competency-based questions. Expect direct questions on mirror formula, magnification, sign convention and application-based questions drawn from NCERT activities. Students who follow NCERT thoroughly and practice this chapter's questions consistently score in the 90%+ range.
What are the key terms to remember for mirror formula & magnification in Class 10 Science?
The key terms to remember for mirror formula & magnification in NCERT Class 10 Science Chapter 9 are: mirror formula, magnification, sign convention, focal length, object distance, image distance. Each of these concepts carries exam weightage and regularly appears in the CBSE board paper. Write clear one-line definitions of every term in your revision notes and revisit them before the exam. Linking these terms visually through a flowchart or concept map makes recall easier during the Class 10 Science board exam.
Is Mirror Formula & Magnification included in the Class 10 Science syllabus for 2025–26 CBSE board exam?
Yes, Mirror Formula & Magnification is a part of the NCERT Class 10 Science syllabus (2025–26) prescribed by CBSE. It falls under Chapter 9 — Light - Reflection and Refraction — and is examined in the annual board paper. The current syllabus retains the full treatment of mirror formula, magnification, sign convention as per the NCERT textbook. Because CBSE bases every board question on NCERT, studying this part thoroughly ensures complete syllabus coverage and guarantees marks from this chapter.
How should I prepare mirror formula & magnification for the CBSE Class 10 Science board exam?
Prepare mirror formula & magnification for the CBSE Class 10 Science board exam in three steps. First, read this NCERT part carefully, highlighting definitions and diagrams of mirror formula, magnification, sign convention. Second, solve every in-text question and end-of-chapter exercise — CBSE questions often come directly from NCERT. Third, practice competency-based and assertion-reason questions to sharpen reasoning. Write answers in the exam-style format (point-wise with diagrams) and time yourself. This method delivers confidence and full marks in the board exam.
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