Assertion (A): A concave lens always forms a virtual and erect image. Reason (R): A concave lens is a converging lens.

Both A and R are true and R is the correct explanation of A.

Both A and R are true but R is not the correct explanation of A.

Explain with the help of a ray diagram why a stick partly immersed in water appears to be bent at the water surface.

It is due to refraction. Light rays from the immersed part travel from water (denser) to air (rarer), bending away from the normal. The emergent rays appear to come from a point above the actual position, making the stick look bent. A ray diagram would show incident rays from underwater bending away from the normal at the interface, and when traced backwards, they meet at a higher point.

A ray of light travelling in air enters obliquely into a glass slab. Does it bend towards or away from the normal? Explain why.

It bends towards the normal because glass is optically denser than air, light slows down, and a ray passing from a rarer to a denser medium bends towards the normal.

The refractive index of glass with respect to air is 3/2 and that of water with respect to air is 4/3. If the speed of light in glass is 2 × 10^8 m/s, calculate the speed of light in water and the refractive index of water with respect to glass.

Speed of light in water = 2.25 × 10^8 m/s; refractive index of water with respect to glass = 8/9.

State the laws of reflection of light. With the help of a suitable ray diagram, explain the formation of an image by a plane mirror. List four characteristics of the image formed.

(a) Laws of reflection: (i) The incident ray, the reflected ray, and the normal to the reflecting surface at the point of incidence all lie in the same plane. (ii) The angle of incidence is equal to the angle of reflection, i.e., ∠i = ∠r. (b) Image formation by a plane mirror: When a point object O is placed in front of a plane mirror, take two rays OA and OB from O striking the mirror at A and B. The reflected rays obey the law of reflection. When extended backwards, they appear to intersect at I behind the mirror. By geometry (congruent triangles), object distance OA equals image distance IA. Thus, a virtual and erect image is formed at I. (c) Characteristics: (i) Virtual and erect. (ii) Same size as the object. (iii) Laterally inverted. (iv) Image distance behind the mirror equals object distance in front.

With the help of a neat ray diagram, trace the path of a light ray passing through a rectangular glass slab. Show that the emergent ray is parallel to the incident ray but laterally displaced. Derive an expression for the lateral displacement in terms of the angle of incidence i, angle of refraction r, and thickness t of the slab.

Ray diagram: Draw a slab with parallel sides. An incident ray enters at angle i, refracts at angle r, travels through the slab, strikes the second face at angle r, and emerges at angle i again (by reversibility). Thus, incident and emergent rays are parallel. The perpendicular distance between them is the lateral displacement d. In the triangle formed inside the slab (thickness t, angle r), the path length in the slab = t / cos r. The lateral shift is the horizontal component of this path: d = (t / cos r) × sin(i - r). Using trigonometric identities, d = t (sin i cos r - cos i sin r)/cos r = t (sin i - cos i tan r). If i is small, d ≈ t i (1 - 1/n). This shows the shift depends on i, r, and t; it reduces to zero for normal incidence (i = 0).

An object 2 cm tall is placed perpendicular to the principal axis of a convex lens of focal length 15 cm. The distance of the object from the lens is 20 cm. (1) Find the position of the image. (2) Calculate the magnification. (3) State the nature of the image.

Image at 60 cm behind the lens; magnification -3, image 6 cm tall; real, inverted, magnified.

Class 10 Science Chapter 9: Light – Reflection and Refraction — Important Questions & Sample Paper

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Chapter 9 'Light – Reflection and Refraction' is a cornerstone of Class 10 Physics, blending fundamental optics principles with real-world applications. The chapter begins with the laws of reflection and the properties of plane and spherical mirrors. Students learn to construct ray diagrams and apply the mirror formula (1/f = 1/v + 1/u) and magnification (m = -v/u) to concave and convex mirrors, determining image position, size, and nature. The concepts of real, virtual, erect, and inverted images are crucial.

Moving to refraction, the chapter explains how light bends when traveling between media of different densities. Key topics include Snell’s law (n1 sin i = n2 sin r), absolute refractive index, and the behavior of light through glass slabs and prisms. Lenses—both convex and converging—are analyzed using the lens formula (1/f = 1/v - 1/u) and magnification, with emphasis on special rays and image formation.

Exam questions typically include: numerical problems on mirror and lens formulas, sometimes requiring two-step solutions with altered object positions; conceptual queries about image characteristics (e.g., where to place an object for a virtual, erect, magnified image in a concave mirror); drawing ray diagrams for specific cases; and refractive index calculations, including angle of incidence/refraction. A solid grasp of sign conventions and formula application is essential for success.

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Science — Light – Reflection and Refraction

Class 10Time: 3 hrsMax Marks: 80

SECTION A

1.
In the experiment to study the dependence of angle of deviation (D) on angle of incidence (i) for a prism, a graph is plotted with i on the x-axis and D on the y-axis. What is the slope of the graph at the point of minimum deviation?
(a) 0(b) 1(c) -1(d) Infinity
1
2.
A convex lens of focal length 25 cm forms a real image at a distance of 50 cm from the lens. What is the object distance?
(a) -50 cm(b) 50 cm(c) -25 cm(d) 25 cm
1
3.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
(a) 20 cm(b) 10 cm(c) 5 cm(d) 40 cm
1

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In a CBSE exam, this chapter typically contributes questions across the following types. The last column shows how many original questions of each type we have ready in our bank for this chapter:

Question type	Marks each	In our bank
Multiple Choice (MCQ)	1 mark	13
Assertion–Reason	1 mark	6
Short Answer	2 marks	8
Short Answer	3 marks	6
Long Answer	5 marks	5
Case Study	4 marks	6

44 original, exam-style questions in our bank for this chapter — with answers.

Important & sample questions (with answers)

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Q1. In the experiment to study the dependence of angle of deviation (D) on angle of incidence (i) for a prism, a graph is plotted with i on the x-axis and D on the y-axis. What is the slope of the graph at the point of minimum deviation?
1 mark
Multiple Choice (MCQ)
(A) 0(B) 1(C) -1(D) Infinity
▸ Answer▾ Answer
0
Q2. A convex lens of focal length 25 cm forms a real image at a distance of 50 cm from the lens. What is the object distance?
1 mark
Multiple Choice (MCQ)
(A) -50 cm(B) 50 cm(C) -25 cm(D) 25 cm
▸ Answer▾ Answer
-50 cm
Q3. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
1 mark
Multiple Choice (MCQ)
(A) 20 cm(B) 10 cm(C) 5 cm(D) 40 cm
▸ Answer▾ Answer
10 cm
Q4. A ray of light passing through the center of curvature of a concave mirror strikes the mirror at what angle of incidence?
1 mark
Multiple Choice (MCQ)
(A) 0°(B) 45°(C) 60°(D) 90°
▸ Answer▾ Answer
0°
Q5. Assertion (A): A concave lens always forms a virtual and erect image. Reason (R): A concave lens is a converging lens.
1 mark
Assertion–Reason
(A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.
▸ Answer▾ Answer
A is true but R is false.
Q6. Explain with the help of a ray diagram why a stick partly immersed in water appears to be bent at the water surface.
2 marks
Short Answer
▸ Answer▾ Answer
It is due to refraction. Light rays from the immersed part travel from water (denser) to air (rarer), bending away from the normal. The emergent rays appear to come from a point above the actual position, making the stick look bent. A ray diagram would show incident rays from underwater bending away from the normal at the interface, and when traced backwards, they meet at a higher point.
Q7. A ray of light travelling in air enters obliquely into a glass slab. Does it bend towards or away from the normal? Explain why.
2 marks
Short Answer
▸ Answer▾ Answer
It bends towards the normal because glass is optically denser than air, light slows down, and a ray passing from a rarer to a denser medium bends towards the normal.
Q8. An object is placed 12 cm in front of a convex mirror. Its virtual image is formed 6 cm behind the mirror. Calculate the focal length of the mirror using the mirror formula.
3 marks
Short Answer
▸ Answer▾ Answer
The focal length is 12 cm.
Q9. The refractive index of glass with respect to air is 3/2 and that of water with respect to air is 4/3. If the speed of light in glass is 2 × 10^8 m/s, calculate the speed of light in water and the refractive index of water with respect to glass.
3 marks
Short Answer
▸ Answer▾ Answer
Speed of light in water = 2.25 × 10^8 m/s; refractive index of water with respect to glass = 8/9.
Q10. State the laws of reflection of light. With the help of a suitable ray diagram, explain the formation of an image by a plane mirror. List four characteristics of the image formed.
5 marks
Long Answer
▸ Answer▾ Answer
(a) Laws of reflection: (i) The incident ray, the reflected ray, and the normal to the reflecting surface at the point of incidence all lie in the same plane. (ii) The angle of incidence is equal to the angle of reflection, i.e., ∠i = ∠r. (b) Image formation by a plane mirror: When a point object O is placed in front of a plane mirror, take two rays OA and OB from O striking the mirror at A and B. The reflected rays obey the law of reflection. When extended backwards, they appear to intersect at I behind the mirror. By geometry (congruent triangles), object distance OA equals image distance IA. Thus, a virtual and erect image is formed at I. (c) Characteristics: (i) Virtual and erect. (ii) Same size as the object. (iii) Laterally inverted. (iv) Image distance behind the mirror equals object distance in front.
Q11. With the help of a neat ray diagram, trace the path of a light ray passing through a rectangular glass slab. Show that the emergent ray is parallel to the incident ray but laterally displaced. Derive an expression for the lateral displacement in terms of the angle of incidence i, angle of refraction r, and thickness t of the slab.
5 marks
Long Answer
▸ Answer▾ Answer
Ray diagram: Draw a slab with parallel sides. An incident ray enters at angle i, refracts at angle r, travels through the slab, strikes the second face at angle r, and emerges at angle i again (by reversibility). Thus, incident and emergent rays are parallel. The perpendicular distance between them is the lateral displacement d. In the triangle formed inside the slab (thickness t, angle r), the path length in the slab = t / cos r. The lateral shift is the horizontal component of this path: d = (t / cos r) × sin(i - r). Using trigonometric identities, d = t (sin i cos r - cos i sin r)/cos r = t (sin i - cos i tan r). If i is small, d ≈ t i (1 - 1/n). This shows the shift depends on i, r, and t; it reduces to zero for normal incidence (i = 0).
Q12. An object 2 cm tall is placed perpendicular to the principal axis of a convex lens of focal length 15 cm. The distance of the object from the lens is 20 cm.
4 marks
Case Study
1. (i) Find the position of the image.2 marks
2. (ii) Calculate the magnification.1 mark
3. (iii) State the nature of the image.1 mark
▸ Answer▾ Answer
Image at 60 cm behind the lens; magnification -3, image 6 cm tall; real, inverted, magnified.

Frequently asked questions

What are the important formulas in Light – Reflection and Refraction?

Key formulas include the mirror formula (1/f = 1/v + 1/u), magnification for mirrors (m = -v/u), lens formula (1/f = 1/v - 1/u), magnification for lenses (m = v/u), Snell’s law (n1 sin i = n2 sin r), and absolute refractive index (n = c/v).

How do sign conventions differ for mirrors and lenses in CBSE?

For mirrors: concave mirror focal length (f) and object distance (u) are negative; image distance (v) is negative for real images, positive for virtual. For convex mirrors, f and v are positive, u negative. For lenses: convex lens f positive, concave f negative; u is always negative for real objects; v positive for real images (opposite side), negative for virtual images (same side). All distances measured from pole/optical center.

Which ray diagrams are most likely to appear in Class 10 board exams?

Commonly asked diagrams: concave mirror with object between pole (P) and focus (F) producing a virtual, erect, magnified image; convex lens with object at 2F1 producing a real, inverted, same-size image; and refraction through a glass slab showing lateral displacement. Always label parts and use proper arrow lines.

How to approach numerical problems where object position changes and magnification changes?

First, note all given values with proper signs. Use m = v/u (or -v/u) to express v in terms of u and m. Substitute into the mirror/lens formula. For two magnifications at different positions, set up two equations and solve for f and u. Eliminate variables by equating expressions or using simultaneous equations.

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