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NCERT's Table 9.1 lists six object positions for a concave mirror and the image each one gives. Many students learn it as a block of words, then mix up rows under pressure. This post puts numbers to every row using one mirror, so you can see the pattern instead of memorising it, and then shows how to run the table backwards when a question gives the image you need.
Where is this in the syllabus?
It is in Chapter 9, Light: Reflection and Refraction, of the NCERT Class 10 Science textbook (2026-27 reprint). The CBSE Secondary curriculum for 2026-27 lists "Images formed by spherical mirrors, centre of curvature, principal axis, principal focus, focal length, mirror formula (Derivation not required), magnification". The sign convention and a lens example are in our Class 10 light chapter guide.
What are the six positions?
From NCERT Table 9.1, with P the pole, F the focus and C the centre of curvature:
At infinity: image at F, highly diminished, real and inverted.
Beyond C: image between F and C, diminished, real and inverted.
At C: image at C, same size, real and inverted.
Between C and F: image beyond C, enlarged, real and inverted.
At F: image at infinity. For size and nature, the current table simply says "Image would not be formed".
Between F and P: image behind the mirror, enlarged, virtual and erect.
Worked example: one mirror, all six rows
Take a concave mirror with focal length 12 cm, so its radius of curvature is 24 cm. Use the New Cartesian sign convention: f = −12 cm, C is 24 cm in front, and object distances are negative. The mirror formula is 1/v + 1/u = 1/f, and magnification m = −v/u.
At infinity. 1/u = 0, so v = f = −12 cm. The image is at F.
Beyond C, u = −36 cm. 1/v = −1/12 + 1/36 = −2/36, so v = −18 cm. m = −(−18)/(−36) = −0.5. Real, inverted, half size, between F (12 cm) and C (24 cm).
At C, u = −24 cm. 1/v = −1/12 + 1/24 = −1/24, so v = −24 cm. m = −1. Real, inverted, same size, at C.
Between C and F, u = −18 cm. 1/v = −1/12 + 1/18 = −1/36, so v = −36 cm. m = −2. Real, inverted, twice the size, beyond C.
At F, u = −12 cm. 1/v = −1/12 + 1/12 = 0. The reflected rays are parallel, so the image is at infinity.
Between F and P, u = −6 cm. 1/v = −1/12 + 1/6 = 1/12, so v = +12 cm. m = −12/(−6) = +2. Positive v means behind the mirror. Virtual, erect, twice the size.
Read rows 2 to 4 together. As the object moves in from 36 cm to 18 cm, the image moves out from 18 cm to 36 cm and grows from half size to double size. Rows 2 and 4 are the same pair of points swapped, because light paths are reversible.
How do you run the table backwards?
Exams often give the image and ask for the object position. Use m to link v and u.
Real image three times the size. Real means inverted, so m = −3, and v = 3u. Substitute: 1/(3u) + 1/u = 4/(3u) = −1/12, so u = −16 cm and v = −48 cm. The object is between F (12 cm) and C (24 cm), as row 4 predicts.
Virtual image three times the size. Virtual means erect, so m = +3, and v = −3u. Then 1/(−3u) + 1/u = 2/(3u) = −1/12, so u = −8 cm and v = +24 cm. The object is between F and P, as row 6 predicts.
How do the rows connect to everyday uses?
NCERT lists uses that match particular rows:
Torches, searchlights and headlights: the bulb sits at F (row 5), so the light leaves as a parallel beam.
Shaving mirrors and dentists' mirrors: the face or tooth is between F and P (row 6), giving an enlarged, erect image.
Solar furnaces: sunlight comes from effectively infinity (row 1) and is concentrated at F to produce heat.
Why do students mix up the rows?
Research on how students understand optics suggests the trouble runs deeper than memory. Galili and Hazan (2000) studied high school and teacher-training students before and after ordinary instruction and described stable patterns of alternative ideas about light and images. They recommended teaching to those underlying patterns instead of correcting misconceptions one at a time. For you, that means tying each row to a ray diagram, not to a phrase.
Common slips:
Inventing a size for row 5. The current NCERT table gives the position as infinity and says an image would not be formed. The reflected rays leave parallel and never meet at a finite distance, so learn the row as it is printed in your edition.
Giving row 6 a negative v. A virtual image behind the mirror has positive v.
Using m = v/u. For mirrors, m = −v/u.
Ray diagrams without arrows. Use two rays: one parallel to the axis that reflects through F, and one through C that returns along itself.
A practice routine
Cover this page and draw ray diagrams for all six rows using a mirror with f = 3 cm on your page.
Pick a new focal length, say 15 cm, and calculate v and m for objects at 45 cm, 30 cm, 20 cm and 10 cm. Check that each answer matches a row.
Make up one backwards problem for a real image and one for a virtual image, and solve them.
When you are ready for the lab version of this chapter, see how to find a concave mirror's focal length by experiment.
For seeing how the image moves as you slide the object, the Fermi Simulation Lab has interactive simulations for ideas easier to see than read.
Today, redo the worked example with f = 15 cm and check that each image lands in the row Table 9.1 predicts.
