Class 10 Science · Original EZT practice
Light: Reflection and Refraction (Class 10 Chapter 9): summary and practice
Mirrors and lenses follow a small set of formulas, and almost every numerical in this chapter is those formulas plus careful signs. Chapter 9 covers reflection by spherical mirrors, refraction through glass, lenses and power. All questions are original EZT tutor practice following the chapter's own section order.
8 original worked examples, 8 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 20 October 2026:
1. Reflection and spherical mirrors
Concave mirrors converge light and convex mirrors diverge it. The focal length is half the radius of curvature: R = 2f. Where the object sits - beyond the centre, at the centre, between focus and pole - decides whether the image is real or virtual, inverted or erect, enlarged or diminished.
2. Mirror formula and magnification
The mirror formula is 1/v + 1/u = 1/f, and magnification is m = -v/u. Distances are measured from the pole; distances against the incident light are negative. A negative m means a real, inverted image; a positive m means a virtual, erect one.
3. Refraction and refractive index
Light bends when it enters a new medium because its speed changes. The refractive index is n = c/v, with c = 3 x 10^8 m/s. Through a rectangular glass slab the emergent ray is parallel to the incident ray, shifted sideways.
4. Lenses and power
The lens formula is 1/v - 1/u = 1/f, and magnification is m = v/u. Power is P = 1/f with f in metres; its unit is the dioptre (D). Convex lenses have positive power, concave lenses negative.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Object at the centre of curvature of a concave mirror
Choose the method: Apply the mirror formula with negative f and u, then read m.
EZT practice question
A concave mirror has f = -15 cm. An object sits at u = -30 cm. Find the image position and magnification.
Show the worked solution
- 1/v = 1/f - 1/u = -1/15 + 1/30 = -1/30, so v = -30 cm.
- m = -v/u = -(-30)/(-30) = -1.
Answer: v = -30 cm and m = -1: a real, inverted image of the same size at the centre of curvature.
Check: An object at the centre of curvature always images onto itself with m = -1.
Watch for this: Do not drop the minus signs; f and u are both negative for a concave mirror.
Now try it yourself
For the same mirror, where is the image if u = -45 cm?
Check your answer
v = -22.5 cm.
1/v = -1/15 + 1/45 = -2/45, so v = -45/2 = -22.5 cm.
Object between pole and focus
Choose the method: Inside the focus a concave mirror gives a virtual, erect, enlarged image.
EZT practice question
For the same mirror (f = -15 cm), the object moves to u = -10 cm. Find v and m.
Show the worked solution
- 1/v = 1/f - 1/u = -1/15 + 1/10 = 1/30, so v = +30 cm.
- m = -v/u = -(+30)/(-10) = +3.
Answer: v = +30 cm (behind the mirror, virtual) and m = +3 (erect, enlarged).
Check: A positive v for a mirror means the image is behind it, hence virtual.
Watch for this: Do not read a positive v as a real image; for mirrors, real images form in front.
Now try it yourself
For f = -20 cm and u = -40 cm, find v and m.
Check your answer
v = -40 cm and m = -1.
The object is at the centre of curvature, so it images onto itself, inverted and same size.
Object beyond 2F of a convex lens
Choose the method: Use 1/v - 1/u = 1/f with positive f and negative u.
EZT practice question
A convex lens has f = +20 cm. An object sits at u = -60 cm. Find the image and magnification.
Show the worked solution
- 1/v = 1/f + 1/u = 1/20 - 1/60 = 2/60, so v = +30 cm.
- m = v/u = 30/(-60) = -1/2.
Answer: v = +30 cm and m = -1/2: a real, inverted, diminished image between F and 2F on the other side.
Check: An object beyond 2F images between F and 2F, diminished - that matches.
Watch for this: The lens formula has a minus before 1/u; using the mirror formula here flips the answer.
Now try it yourself
For f = +10 cm and u = -30 cm, find v.
Check your answer
v = +15 cm.
1/v = 1/10 - 1/30 = 2/30, so v = 15 cm.
Convert focal length to power
Choose the method: P = 1/f with f in metres; the sign follows the lens type.
EZT practice question
Find the power of lenses with f = +25 cm, f = -50 cm and f = 20 cm.
Show the worked solution
- Convert to metres: 0.25 m, -0.50 m and 0.20 m.
- P = 1/f for each.
Answer: +4 D, -2 D and +5 D.
Check: Shorter focal length means larger power - 20 cm giving +5 D beats 25 cm giving +4 D.
Watch for this: Do not plug centimetres into P = 1/f; convert to metres first.
Now try it yourself
A lens has power +2.5 D. Find its focal length.
Check your answer
f = 40 cm.
f = 1/P = 1/2.5 = 0.4 m = 40 cm.
Speed of light in a medium
Choose the method: n = c/v with c = 3 x 10^8 m/s.
EZT practice question
The refractive index of glass is 1.5. Find the speed of light in glass.
Show the worked solution
- v = c/n.
- v = 3 x 10^8 / 1.5.
Answer: 2 x 10^8 m/s.
Check: Light slows in an optically denser medium, and 2 x 10^8 is less than 3 x 10^8.
Watch for this: Do not multiply; dividing by n slows the light, multiplying would speed it up.
Now try it yourself
Find the speed of light in a medium with n = 2.
Check your answer
1.5 x 10^8 m/s.
v = 3 x 10^8 / 2.
Apply the sign convention
Choose the method: Measure from the pole or optical centre; against the incident light is negative.
EZT practice question
An object is placed 30 cm in front of a concave mirror. What sign does u carry, and why?
Show the worked solution
- Distances are measured from the pole of the mirror.
- The object lies against the direction of the incident light.
Answer: u = -30 cm; distances measured against the incident light are negative.
Check: Heights above the principal axis are positive, below it negative.
Watch for this: Do not mix conventions between mirrors and lenses mid-problem; set the convention first.
Now try it yourself
What is the sign of the focal length of a convex lens?
Check your answer
Positive.
Its focus lies along the direction of the refracted light.
Read the magnification
Choose the method: The sign of m gives orientation; the size of m gives scaling.
EZT practice question
A mirror forms an image with m = -1. Describe the image fully.
Show the worked solution
- m is negative, so the image is inverted.
- For a mirror, an inverted image is real.
- The size of m is 1, so image and object are the same size.
Answer: Real, inverted and the same size as the object.
Check: A concave mirror gives m = -1 only when the object is at the centre of curvature.
Watch for this: Do not call an m = +1 image real; for mirrors, positive m means virtual and erect.
Now try it yourself
A lens gives m = -2. Describe the image.
Check your answer
Real, inverted and twice the object size.
For a lens, a negative m means a real inverted image; the size doubles.
Relate radius of curvature and focal length
Choose the method: R = 2f for spherical mirrors.
EZT practice question
A concave mirror has a radius of curvature of 40 cm. Find its focal length.
Show the worked solution
- f = R/2.
- f = 40/2 = 20 cm, negative for a concave mirror.
Answer: f = -20 cm.
Check: The focus lies halfway between the pole and the centre of curvature.
Watch for this: Do not forget the sign; concave mirrors take negative f.
Now try it yourself
A convex mirror has f = +15 cm. Find R.
Check your answer
R = 30 cm.
R = 2f = 2 x 15.
Try a mixed set
Choose a method or explanation before opening each answer.
1. State the mirror formula.
Show answer and reason
1/v + 1/u = 1/f.
It links image distance, object distance and focal length for spherical mirrors.
2. State the lens formula.
Show answer and reason
1/v - 1/u = 1/f.
The minus before 1/u is what separates it from the mirror formula.
3. Define power of a lens.
Show answer and reason
P = 1/f with f in metres, measured in dioptres (D).
Opticians prescribe lenses in dioptres, not centimetres.
4. Define refractive index.
Show answer and reason
n = c/v, the speed of light in vacuum over its speed in the medium.
It tells you how much the medium slows light.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- A concave mirror has f = -20 cm and the object is at u = -40 cm. Find v and m.
- A lens has power +2.5 D. Find its focal length in centimetres.
- Find the speed of light in a medium with refractive index 2.
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Questions about this practice page
- How does the sign convention work for mirrors and lenses?
- Distances are measured from the pole or optical centre. Distances against the incident light are negative; heights above the principal axis are positive.
- Why is v positive for an image behind a mirror?
- Behind the mirror is along the direction of the incident light's forward travel, so the distance is positive - and the image is virtual.
- What does a negative magnification mean?
- The image is inverted relative to the object, and for a single mirror or lens that also means it is real.
- What unit is lens power measured in?
- Dioptres (D). Power is the reciprocal of the focal length in metres.
- Are these the NCERT exercise questions for Light - Reflection and Refraction (Class 10)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.