Question 1 · MCQ · Analyse · 1 mark
A wide rear view
Spherical mirrors
Why is a convex mirror useful as a vehicle rear-view mirror?
(A) It always forms enlarged real images.
(B) It gives erect diminished images and a wide field of view.
(C) It forms images only at its pole.
(D) It has no principal focus.
Show answer and solution
Answer: (B) It gives erect diminished images and a wide field of view.
Worked solution: A convex mirror forms an erect, virtual and diminished image for real objects in front of it. Its wide field of view helps the driver see more area behind.
Question 2 · MCQ · Analyse · 1 mark
A magnifier close to a page
Lenses and image formation
A convex lens is held with the printed page inside its focal length. What image is seen through the lens?
(A) Real, inverted and diminished
(B) Real and on a distant screen
(C) Virtual, erect and enlarged
(D) Virtual, inverted and unchanged in size
Show answer and solution
Answer: (C) Virtual, erect and enlarged
Worked solution: For an object between a convex lens and its focus, the refracted rays diverge and appear to come from an enlarged erect virtual image on the object side.
Question 3 · MCQ · Analyse · 1 mark
Reading a lens label
Power of a lens
A converging lens has focal length +0.50 m. What is its power?
(A) +2 D
(B) −2 D
(C) +0.5 D
(D) −0.5 D
Show answer and solution
Answer: (A) +2 D
Worked solution: Power P = 1/f when f is in metres. P = 1/0.50 = +2 dioptres; the positive sign identifies a converging lens.
Question 4 · MCQ · Analyse · 1 mark
A ray enters glass
Refraction and refractive index
A light ray enters glass from air obliquely. Which path is expected at the first surface?
(A) It bends away from the normal in every case.
(B) It stops at the surface.
(C) It travels along the normal regardless of incidence.
(D) It bends towards the normal.
Show answer and solution
Answer: (D) It bends towards the normal.
Worked solution: Glass has a higher refractive index than air. Light slows down and bends towards the normal when entering obliquely from air.
Question 5 · A-R · Analyse · 1 mark
A displaced ray in a slab
Refraction and refractive index
Assertion (A): A ray emerging from a rectangular glass slab into air can be parallel to the incident ray but laterally displaced.
Reason (R): The two refracting faces are parallel, so the angular deviation at the second face compensates for that at the first.
(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not correctly explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show answer and solution
Answer: (A) Both A and R are true, and R correctly explains A.
Worked solution: For parallel faces with air on both sides, the emergent angle equals the incident angle. The ray has nevertheless travelled along a different path inside the slab, producing lateral displacement for oblique incidence.
Question 6 · A-R · Analyse · 1 mark
A real image and a screen
Spherical mirrors
Assertion (A): A virtual image formed by a convex mirror can be caught directly on a screen placed at that image position.
Reason (R): Reflected rays do not actually meet at the apparent position of a virtual image.
(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not correctly explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show answer and solution
Answer: (D) A is false, but R is true.
Worked solution: A screen forms an image where light rays actually converge. A virtual image is an apparent intersection of backward extensions, so it cannot be projected directly there.
Question 7 · VSA · Apply · 2 marks
Finding focus from curvature
Spherical mirrors
A concave mirror has radius of curvature 30 cm. State its focal length using the Cartesian sign convention for a mirror facing the object.
Show answer and solution
Answer: f = −15 cm.
Worked solution: The magnitude of focal length is half the radius: 30/2 = 15 cm. The focus of a concave mirror is in front of the mirror, so f is negative in this convention.
Question 8 · VSA · Apply · 2 marks
Light speed in glass
Refraction and refractive index
The speed of light in vacuum is 3.0 × 10⁸ m/s and a glass sample has refractive index 1.5. Find the speed of light in the glass.
Show answer and solution
Answer: 2.0 × 10⁸ m/s.
Worked solution: Refractive index n = c/v, so v = c/n = (3.0 × 10⁸)/1.5 = 2.0 × 10⁸ m/s.
Question 9 · SA · Analyse · 3 marks
A concave mirror forms an image
Spherical mirrors
An object is 30 cm in front of a concave mirror of focal length 15 cm. Using the Cartesian sign convention, calculate image distance and magnification, and describe the image.
Show answer and solution
Answer: v = −30 cm; m = −1; real, inverted and the same size.
Worked solution: Use u = −30 cm and f = −15 cm. From 1/f = 1/v + 1/u, 1/v = −1/15 + 1/30 = −1/30. Thus v = −30 cm. Magnification m = −v/u = −1, indicating an inverted image of equal size.
Question 10 · SA · Analyse · 3 marks
A converging lens makes a real image
Lenses and image formation
An object is placed 30 cm to the left of a convex lens of focal length 20 cm. Find the image distance and magnification using the Cartesian sign convention.
Show answer and solution
Answer: v = +60 cm; m = −2; the image is real, inverted and twice the object size.
Worked solution: Here u = −30 cm and f = +20 cm. The lens formula is 1/f = 1/v − 1/u, so 1/v = 1/20 − 1/30 = 1/60. Thus v = +60 cm and m = v/u = −2.
Question 11 · LA · Analyse · 5 marks
Tracking a concave-lens image
Lenses and image formation
An object of height 6 cm is 30 cm to the left of a concave lens with focal length −15 cm. Calculate (i) image distance, (ii) magnification, (iii) image height, (iv) power in dioptres, and (v) state the image nature.
Show answer and solution
Answer: v = −10 cm; m = +1/3; image height = +2 cm; P ≈ −6.67 D; virtual, erect and diminished.
Worked solution: With u = −30 cm and f = −15 cm, 1/v = 1/f + 1/u = −1/15 − 1/30 = −1/10. Thus v = −10 cm. m = v/u = 1/3 and image height = (1/3) × 6 = +2 cm. P = 1/(−0.15) ≈ −6.67 D. Positive magnification and negative image distance indicate an erect virtual image on the object side.
Question 12 · Case Study · Analyse · 4 marks
Moving an object near a convex lens
Lenses and image formation
A convex lens has focal length 10 cm. In trial P an object is 30 cm to its left; in trial Q the object is 5 cm to its left. Use the Cartesian sign convention.
Answer the following:
(i) Calculate the image distance in trial P. [1 mark]
(ii) Can the image in trial P be caught on a screen? [1 mark]
(iii) Calculate the image distance in trial Q and describe its nature. [2 marks]
Show answer and solution
Answer: (i) +15 cm. (ii) Yes. (iii) −10 cm; virtual, erect and enlarged.
Worked solution: For P, 1/v = 1/10 − 1/30 = 1/15, so a real image forms 15 cm to the right. For Q, 1/v = 1/10 − 1/5 = −1/10, so v = −10 cm. The magnification in Q is (−10)/(−5) = +2, giving an erect enlarged virtual image.
