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Topic 8 Practice Paper: Light — mark scheme

QuizQuipPractice Series
QQ/IGCSE-PHYS/T8

IGCSE · Cambridge IGCSE Physics (Extended (Core + Supplement))

Mark scheme

Light

Topic 8 Practice Paper · Maximum mark: 30

Notes for markers

  • Award one mark for each point marked [1], up to the marks shown for the part.
  • Accept answers worded differently that show the same understanding.
  • Units are required where the answer line shows one. ecf: allow an error carried forward.

Section A: Multiple-choice questions[10 marks]

QuestionAnswer
1AFrom the surface, convert first: angle of incidence = 90° − 20° = 70°, so reflection is also 70°.1
2CPlane mirror: virtual, upright, same size, as far behind as the object is in front.1
3BSpeeding up on leaving the glass bends the ray away from the normal.1
4DInside the denser medium at more than the critical angle: total internal reflection, no light escapes.1
5An = 1/sin c = 1 ÷ sin 45° = 1 ÷ 0.707 = 1.41.1
6CEach wall strike is beyond the critical angle, so the light is trapped and guided by total internal reflection.1
7BDiverging lenses spread parallel rays outward; extrapolated back, they appear to originate from a virtual focus in front of the lens.1
8DInside the focal length the refracted rays diverge — the magnifying-glass arrangement: upright, enlarged, virtual.1
9ARed travels fastest in glass, so it bends the least; violet bends the most.1
10BAlong the normal there is no angle to change: light enters straight on and keeps going straight.1

Section B: Structured questions[20 marks]

QuestionAnswer and marking pointsMarks
11(a)

35°: the angle of reflection equals the angle of incidence, both measured from the normal.

  • 35° [1]
  • law of reflection quoted [1]
2
11(b)

6.0 cm behind the mirror, on the normal line through the torch. [1]

1
11(c)

The reflected rays diverge, and extrapolating them backwards they appear to meet 6.0 cm behind the mirror — a virtual image.

  • rays appear to come from behind the mirror [1]
  • virtual image named [1]
2
Total for Question 115
12(a)

It bends towards the normal, because it slows down entering the optically denser glass.

  • towards the normal [1]
1
12(b)

n = sin 40° ÷ sin 25° = 0.643 ÷ 0.423 = 1.52.

  • n = sin i / sin r substituted [1]
  • 1.52 [1]
2
12(c)

sin c = 1 ÷ 1.52 = 0.658, so c = sin⁻¹(0.658) ≈ 41°.

  • sin c = 1/n [1]
  • ≈ 41° [1]
2
12(d)

The light must be travelling inside the denser medium (the glass), towards the less dense air.

  • travel in the denser medium [1]
1
Total for Question 126
13(a)

Just under 5.0 cm — the stamp must sit inside the focal length.

  • inside the focal length (< 5.0 cm) [1]
1
13(b)

The image is enlarged, upright and virtual.

  • enlarged [1]
  • upright [1]
  • virtual [1]
3
13(c)

The rays leaving the lens are diverging and only appear to come from the image position, so no image can be caught on a screen there.

  • diverging rays extrapolated back / cannot form on a screen [1]
1
Total for Question 135
14(a)

Dispersion: each colour travels at a different speed in glass, so each is refracted by a different amount.

  • dispersion [1]
  • different speeds/refractions for different colours [1]
2
14(b)

Red at the top (bends least); violet at the bottom (bends most).

  • red least, violet most [1]
1
14(c)

Only a single green band — no spectrum; the light is monochromatic.

  • single green band / monochromatic [1]
1
Total for Question 144

— End of mark scheme —