Class 10 Science Case Study Worksheet Chapter 9 Light Reflection and Refraction | Detailed Solutions & Marking Scheme

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 VEDANT WORKSHEET SERIES

Class: 10
Subject: SCIENCE
Chapter: Light – Reflection and Refraction
Time: 1 Hour
Maximum Marks: 20

Section A: Case Study Based Questions (5 × 4 = 20 Marks)

Case Study 1:
Riya is performing an experiment using a concave mirror. She places an object beyond the centre of curvature and observes that the image formed is real, inverted, and diminished. She then gradually moves the object towards the mirror and observes changes in the image position and size.

Questions:
a) Where is the image formed when the object is at the centre of curvature? (1)
b) What happens to the image when the object is placed between F and C? (1)
c) Using mirror formula, find the focal length if object distance u = -30 cm and image distance v = -15 cm. (2)

Case Study 2:
A student passes a ray of light from air into a glass slab. He observes that the ray bends towards the normal when entering the glass and bends away from the normal when emerging into air.

Questions:
a) State Snell’s law of refraction. (1)
b) What is refractive index? (1)
c) If angle of incidence i = 30° and angle of refraction r = 20°, calculate refractive index. (2)

Case Study 3:
A convex lens is used to project the image of a candle flame on a screen. The image is found to be real and inverted. When the lens is moved farther from the screen, the image becomes blurred.

Questions:
a) Why is the image real in this case? (1)
b) What happens to image size when object is moved closer to lens? (1)
c) Calculate focal length if u = -20 cm and v = 40 cm using lens formula:
1/f = 1/v + 1/u (2)

Case Study 4:
A student studies the magnification produced by a mirror. He finds that when the image is erect and virtual, the magnification is positive, while for inverted images, magnification is negative.

Questions:
a) Write the formula for magnification for mirrors. (1)
b) What does positive magnification indicate? (1)
c) Find magnification if image height hi = 6 cm and object height ho = 3 cm. Also state nature of image. (2)

Case Study 5:
A person uses spectacles with power -2 D for correcting vision. The lens used helps in proper focusing of light on the retina.

Questions:
a) What type of lens is used in spectacles of negative power? (1)
b) Define power of a lens. (1)
c) Calculate focal length of the lens using P = 1/f. (2)

VEDANT WORKSHEET SERIES - ANSWER KEY WITH MARKING SCHEME
Class: 10 | Subject: SCIENCE
Chapter: Light – Reflection and Refraction

Section A: Case Study Based Questions

Case Study 1

a) Image is formed at the centre of curvature (C).
Marks: 1

b) Image formed is real, inverted, and enlarged, formed beyond C.
Marks: 1

c) Mirror formula:
1/f = 1/v + 1/u

Substitute values:
u = -30 cm, v = -15 cm

1/f = (1/(-15)) + (1/(-30))
1/f = -2/30 - 1/30
1/f = -3/30
1/f = -1/10

f = -10 cm

Answer: f = -10 cm
Marks: 2 (1 for formula + substitution, 1 for correct answer)

Case Study 2

a) Snell’s Law:
n = sin i / sin r

Marks: 1

b) Refractive index is the ratio of speed of light in vacuum to speed of light in medium OR ratio of sin i to sin r.
Marks: 1

c) Given: i = 30°, r = 20°

n = sin 30° / sin 20°
n = 0.5 / 0.342

n ≈ 1.46

Answer: n ≈ 1.46
Marks: 2 (1 for formula, 1 for calculation)

Case Study 3

a) Image is real because rays actually converge on the screen.
Marks: 1

b) Image size increases when object moves closer to lens (towards F).
Marks: 1

c) Lens formula:
1/f = 1/v + 1/u

u = -20 cm, v = 40 cm

1/f = 1/40 + 1/(-20)
1/f = 1/40 - 2/40
1/f = -1/40

f = -40 cm

Answer: f = -40 cm
Marks: 2 (1 for formula, 1 for correct answer)

Case Study 4

a) Magnification formula:
m = hi / ho

Marks: 1

b) Positive magnification indicates image is virtual and erect.
Marks: 1

c) Given: hi = 6 cm, ho = 3 cm

m = 6 / 3 = +2

Image is erect and magnified (virtual).

Answer: m = +2, virtual and erect image
Marks: 2 (1 for calculation, 1 for interpretation)

Case Study 5

a) Concave lens is used.
Marks: 1

b) Power of lens is defined as reciprocal of focal length (in metres).
P = 1/f

Marks: 1

c) Given: P = -2 D

P = 1/f

-2 = 1/f

f = -1/2 m = -0.5 m

Answer: f = -0.5 m
Marks: 2 (1 for formula, 1 for correct answer)

Total Marks: 20


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