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Unit IX · Light energy (lesson 15)
10 min readConcept sheetSection A MCQ (1 mark) on signs and the formula; Section B short answer (3 marks) numerical on the mirror formula and magnification

Sign convention, mirror formula and magnification

Measure every distance from the pole with the sign convention. Then use the mirror formula 1/f = 1/v + 1/u and the magnification m = −v/u.

Mirror formula1/f = 1/v + 1/u (u: object distance, v: image distance, f: focal length)
Magnificationm = size of image (I) ÷ size of object (O) = −v/u; your NIOS lesson writes v/u, which gives only the size
Sign of fConcave mirror: negative; convex mirror: positive
Signs of v and mv negative, m negative: real and inverted, in front. v positive, m positive: virtual and erect, behind

The sign convention (as in your NIOS lesson)

  • Measure all distances from the pole P of the mirror.
  • Place the object on the left, so light travels from left to right.
  • Distances measured in the direction of the incident light (to the right of P) are positive.
  • Distances measured against the incident light (to the left of P) are negative.
  • Heights above the principal axis are positive; heights below it are negative.
  • So the object distance u is always negative. The focus of a concave mirror is in front, so its f is negative. A convex mirror has a positive f.

Formula, magnification and what the signs tell you

Mirror formula1/f = 1/v + 1/u
To find v1/v = 1/f − 1/u. Put in u and f with their signs.
Magnification mm = size of image (I) ÷ size of object (O) = −v/u. Heights above the axis are positive and heights below are negative.
Size from mIgnore the sign for size: more than 1 means enlarged, less than 1 means smaller, and 1 means the same size.
Nature from the signsv negative and m negative: real and inverted, in front of the mirror. v positive and m positive: virtual and erect, behind the mirror.
Your NIOS lessonIt writes M = I/O = v/u. That gives the right size, but not the right sign. Use m = −v/u for a mirror, so the sign tells you real or virtual. For a lens, m = v/u is correct as it is.

Check the formula with the object at C

  1. 1
    When the object is at C, the image is also at C (see the image table).
  2. 2
    So u = −R and v = −R, where R is the radius of curvature.
  3. 3
    Then 1/v + 1/u = −1/R − 1/R = −2/R.
  4. 4
    For a concave mirror f = −R/2, so 1/f = −2/R.
  5. 5★
    Both sides are equal, so the mirror formula agrees with the ray diagram. It also works for a convex mirror.

★ key step

Concave mirror: a real image

Problem

An object 3 cm tall stands 25 cm in front of a concave mirror of focal length 15 cm. Find the position, size and nature of the image.

Working
  1. Given with signs: u = −25 cm, f = −15 cm (concave), object height O = 3 cm.
  2. 1/v = 1/f − 1/u = 1/(−15) − 1/(−25) = −1/15 + 1/25.
  3. LCM is 75: 1/v = (−5 + 3)/75 = −2/75, so v = −75/2 = −37.5 cm.
  4. m = −v/u = −(−37.5)/(−25) = −1.5.
  5. Image height I = m × O = −1.5 × 3 cm = −4.5 cm. The minus sign means the image is inverted.
Answer

The image is 37.5 cm in front of the mirror (v is negative). It is real, inverted (m is negative) and larger: 4.5 cm tall.

Convex mirror: a scooter's rear-view mirror

Problem

A scooter's convex mirror has a radius of curvature of 2.0 m. A car is 4.0 m behind the scooter, in front of the mirror. Find the position, nature and size of its image.

Working
  1. Given: R = +2.0 m, so f = R/2 = +1.0 m (convex). u = −4.0 m.
  2. 1/v = 1/f − 1/u = 1/1.0 − 1/(−4.0) = 1 + 0.25 = 1.25.
  3. v = 1 ÷ 1.25 = +0.8 m.
  4. m = −v/u = −(0.8)/(−4.0) = +0.2.
Answer

The image is 0.8 m behind the mirror (v is positive). It is virtual, erect (m is positive) and one-fifth the size of the car.

Concave mirror: a make-up mirror

Problem

A face is 10 cm from a concave make-up mirror of focal length 20 cm. Where is the image, and how large is it?

Working
  1. Given: u = −10 cm, f = −20 cm.
  2. 1/v = 1/f − 1/u = −1/20 + 1/10 = (−1 + 2)/20 = 1/20.
  3. v = +20 cm.
  4. m = −v/u = −(20)/(−10) = +2.
Answer

The image is 20 cm behind the mirror. It is virtual, erect (m is positive) and twice the size of the face.

In real life

A shaving or make-up mirror is concave, so a face held inside F looks two or three times bigger. The round security mirror in a shop corner is convex: its image is always small and upright, with v positive.

Exam angle

1 marks: “In the sign convention, what is the sign of the focal length of a concave mirror?” Negative, because its focus lies in front of the mirror, against the direction of the incident light.

2 marks: “A concave mirror forms a virtual image three times the size of the object. Where is the object, and is v positive or negative?” The object is between the pole and the focus. The image is behind the mirror, so v is positive, and m = +3.

3 marks: “An object 2 cm tall is placed 10 cm in front of a concave mirror of focal length 15 cm. Find the position, size and nature of the image.” u = −10 cm, f = −15 cm. 1/v = −1/15 + 1/10 = (−2 + 3)/30 = 1/30, so v = +30 cm (behind). m = −v/u = −(30)/(−10) = +3, so I = 3 × 2 cm = 6 cm. Virtual, erect, enlarged.

Remember it

Left is less: anything to the left of the pole is negative. The object is always on the left, so u is always minus.

Read alongside the books on your syllabus: NIOS, Science and Technology (212): Self Learning Material, Books 1 to 3 · NIOS, Science and Technology (212): Laboratory Manual. Written from standard textbooks; faculty review in progress. Spotted an error? Report it.