Class 10 Science QP Worksheet Chapter 9 Light Reflection and Refraction | Detailed Solutions & Marking Scheme
VEDANT WORKSHEET SERIES
Class: 10
Subject: Science
Chapter: Light – Reflection and Refraction
Time: 1 Hour
Maximum Marks: 30
Section A (1 × 7 = 7 Marks)
Q1. The radius of curvature of a spherical mirror is 20 cm. Its focal length is:
A. 5 cm
B. 10 cm
C. 20 cm
D. 40 cm
Q2. A concave mirror forms a real and inverted image. The object must be placed:
A. Between pole and focus
B. Between focus and centre of curvature
C. At infinity
D. Behind the mirror
Q3. The refractive index of a medium depends on:
A. Colour of light
B. Nature of medium
C. Both A and B
D. Distance travelled
Q4. A ray of light passes from air to glass. The angle of refraction is smaller than angle of incidence because:
A. Glass is rarer medium
B. Glass is denser medium
C. Speed of light increases
D. Frequency changes
Q5. The power of a convex lens is +2 D. Its focal length is:
A. 0.5 m
B. 2 m
C. -0.5 m
D. -2 m
Q6. Assertion (A): A concave mirror can form both real and virtual images.
Reason (R): The nature of image depends on object position.
A. Both A and R are true, R is correct explanation of A
B. Both A and R are true, R is not correct explanation of A
C. A is true, R is false
D. A is false, R is true
Q7. Assertion (A): When light travels from denser to rarer medium, it bends away from normal.
Reason (R): Speed of light increases in rarer medium.
A. Both A and R are true, R is correct explanation of A
B. Both A and R are true, R is not correct explanation of A
C. A is true, R is false
D. A is false, R is true
Section B (2 × 4 = 8 Marks)
Q8. Define principal focus and focal length of a concave mirror.
Q9. State the laws of refraction of light.
Q10. A concave mirror has focal length 15 cm. Find its radius of curvature.
Q11. Differentiate between real and virtual images (any two points).
Section C (3 × 2 = 6 Marks)
Q12. A ray of light enters from air (n = 1) into a medium of refractive index 1.5 at an angle of incidence 30°. Calculate the angle of refraction.
Q13. An object is placed at a distance of 30 cm from a convex lens of focal length 15 cm. Find the position of the image using lens formula.
Section D (5 × 1 = 5 Marks)
Q14. A concave mirror produces an image of magnification -2 when an object is placed at a distance of 20 cm.
(i) Find the image distance.
(ii) Determine focal length of the mirror.
(iii) State the nature of the image.
Section E (Case Study) (4 Marks)
Q15. Read the following and answer the questions:
A student performs an experiment using a convex lens to form images of an object placed at different distances. He observes that when the object is placed beyond 2F, the image is formed between F and 2F, real and inverted. When the object is placed between F and 2F, the image is formed beyond 2F.
(i) What happens to the image when the object is placed at F? (1 mark)
(ii) State the nature of image when object is placed between F and optical centre. (1 mark)
(iii) Draw ray diagram for object placed beyond 2F and describe image formation. (2 marks)
Here is the detailed solved Answer Key with marking scheme (CBSE format, easy to copy, plain text):
VEDANT WORKSHEET SERIES - ANSWER KEY (2026-27)
Chapter: Light – Reflection and Refraction
Section A (1 × 7 = 7 Marks)
Q1. Answer: B (10 cm)
f = R/2 = 20/2 = 10 cm
Marks: 1
Q2. Answer: B (Between focus and centre of curvature)
Real and inverted image is formed when object is beyond F.
Marks: 1
Q3. Answer: C (Both A and B)
Refractive index depends on nature of medium and wavelength (colour).
Marks: 1
Q4. Answer: B (Glass is denser medium)
Light bends towards normal in denser medium due to decrease in speed.
Marks: 1
Q5. Answer: A (0.5 m)
P = 1/f
f = 1/2 = 0.5 m
Marks: 1
Q6. Answer: A
Both statements true and reason correctly explains assertion.
Marks: 1
Q7. Answer: A
Speed increases in rarer medium → bends away from normal.
Marks: 1
Section B (2 × 4 = 8 Marks)
Q8.
Principal Focus: Point where parallel rays meet after reflection. (1 mark)
Focal Length: Distance between pole and principal focus. (1 mark)
Total: 2 Marks
Q9. Laws of Refraction:
Incident ray, refracted ray and normal lie in same plane. (1 mark)
sin i / sin r = constant (refractive index). (1 mark)
Total: 2 Marks
Q10.
R = 2f = 2 × 15 = 30 cm
Marks: 2
Q11. Differences:
Any two points:
Real Image:
• Can be obtained on screen
• Inverted
Virtual Image:
• Cannot be obtained on screen
• Erect
(1 mark each point)
Total: 2 Marks
Section C (3 × 2 = 6 Marks)
Q12.
Given: n = 1.5, i = 30°
Using Snell’s law:
n = sin i / sin r
1.5 = sin30° / sin r
1.5 = 0.5 / sin r
sin r = 0.5 / 1.5 = 1/3
r ≈ 19.5°
Final Answer: 19.5°
Marking Scheme:
Formula: 1 mark
Calculation: 1 mark
Final answer: 1 mark
Total: 3 Marks
Q13.
Given: u = -30 cm, f = +15 cm
Lens formula:
1/f = 1/v - 1/u
1/15 = 1/v + 1/30
1/v = 1/15 - 1/30
1/v = (2 - 1)/30 = 1/30
v = +30 cm
Image formed at 30 cm on other side (real, inverted)
Marking Scheme:
Formula: 1 mark
Substitution: 1 mark
Final answer + nature: 1 mark
Total: 3 Marks
Section D (5 Marks)
Q14.
Given: m = -2, u = -20 cm
(i) m = -v/u
-2 = -v / (-20)
-2 = v/20
v = -40 cm
(2 marks)
(ii) Mirror formula:
1/f = 1/v + 1/u
1/f = 1/(-40) + 1/(-20)
1/f = (-1/40 - 2/40) = -3/40
f = -13.3 cm
(2 marks)
(iii) Nature:
Real, inverted, magnified
(1 mark)
Total: 5 Marks
Section E (Case Study) (4 Marks)
Q15.
(i) Object at F → Image at infinity
Highly enlarged
Marks: 1
(ii) Between F and optical centre →
Image is virtual, erect and magnified
Marks: 1
(iii) Ray diagram explanation:**
• One ray parallel → passes through focus
• One ray through optical centre → undeviated
• Image forms between F and 2F
Nature: real, inverted, diminished
Marking Scheme:
Ray explanation: 1 mark
Image nature + position: 1 mark
Total: 2 Marks
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