3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
A flat, uniformly luminous circular disc of diameter 4.0cm is a light source. An opaque circular disc of diameter 2.0cm is held parallel to the source and coaxial with it, 50cm from it, and a large screen is placed parallel to both and coaxial with them, 150cm from the source (on the far side of the opaque disc). Light travels in straight lines; neglect diffraction and reflections. On the screen, the shadow pattern centred on the axis consists of ______. Exactly one option is correct.
Aa region of total shadow (umbra) of diameter 2.0cm, surrounded by partial shadow out to a diameter of 14cm
Bno region of total shadow; a central disc of diameter 2.0cm from every point of which an annular ring of the source is visible, with partial shadow extending out to a diameter of 14cm
Ca region of total shadow of diameter 6.0cm, surrounded by partial shadow out to a diameter of 18cm
Dno region of total shadow; partial shadow extending out to a diameter of 10cm
Show answer and solution
Answer: B
Take the axis as the x-axis with the source at x=0; source radius R=2cm, obstacle radius r=1cm at x=50cm, screen at x=150cm; measure y in one meridian plane.
Inner boundary (line from the top edge of the source through the top edge of the obstacle): y=2−x/50. It meets the axis at x=100cm, so the umbra cone ends 50cm before the screen. At x=150, y=−1cm: beyond the vertex the cone reopens as the antumbra of radius 1cm.
Check on the axis at the screen: the obstacle (distance 100cm) subtends angular radius 1/100=0.010, the source (distance 150cm) subtends 2/150=0.0133, so a bright ring of the source is visible round the obstacle. No point of the screen is in total shadow.
Outer boundary (top of source through bottom of obstacle): y=2−3x/50; at x=150, y=−7cm, so partial shadow extends to radius 7cm (diameter 14cm).
Trap: reading the inner-boundary value ∣y∣=1cm as an umbra radius (option A) ignores that the lines crossed at x=100cm.
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
In a horizontal xy-plane, a large plane mirror stands vertically. Its reflecting surface lies along a line making 60∘ with the +x axis (measured anticlockwise from +x). The mirror keeps this orientation while the whole mirror translates with velocity 2.0m/s along +y. A small object in front of the mirror moves with velocity 4.0m/s along +x. With i^, j^ the unit vectors along +x, +y, the velocity of the image of the object at this instant is ______. Exactly one option is correct.
A(−2i^+23j^)m/s
B[−(2+23)i^+(23+21)j^]m/s
C[(2+3)i^−(1+23)j^]m/s
D[−(2+3)i^+(1+23)j^]m/s
Show answer and solution
Answer: D
Unit normal to the mirror: n^=(−sin60∘,cos60∘)=(−23,21).
In the mirror's frame the image velocity is the object velocity with its normal component reversed. Back in the ground frame (only the normal component of the mirror's velocity matters):
vI=vO−2[(vO−vM)⋅n^]n^.
vO−vM=(4,−2); (vO−vM)⋅n^=−23−1.
vI=(4,0)+2(1+23)(−23,21)=(4−3−6,1+23)=(−(2+3),1+23)m/s, speed ≈5.82m/s.
Option A ignores the mirror's motion; option B counts the mirror's normal velocity once instead of twice; option C reverses the whole vector.
Difficulty: Very hard
Question 3Level 3 · AdvancedSingle correct
A concave mirror of focal length 20cm has its principal axis along the x-axis, its pole at the origin and its reflecting surface facing the −x side, where the object is. At a certain instant a small point object is at x=−30cm, y=+0.50cm and has velocity (+2.0i^+1.0j^)cm/s. Use paraxial optics. The velocity of the image at this instant is ______. Exactly one option is correct.
A(−8.0i^−2.2j^)cm/s
B(+8.0i^−2.0j^)cm/s
C(−8.0i^−2.0j^)cm/s
D(−8.0i^−1.8j^)cm/s
Show answer and solution
Answer: A
Cartesian convention: u=−30cm, f=−20cm, v1+u1=f1 gives v=u−fuf=−60cm; m=−uv=f−uf=−2.
Axial component: differentiating the mirror equation, dtdv=−u2v2dtdu=−4×2=−8cm/s (image moves away from the mirror).
Transverse component: yI=myO with m=f−uf, so dtdm=(f−u)2fdtdu=100−20×2=−0.4s−1.
dtdyI=mdtdyO+yOdtdm=(−2)(1.0)+(0.50)(−0.4)=−2.2cm/s.
vI=(−8.0i^−2.2j^)cm/s.
Trap: option C drops the yOdm/dt term, which is non-zero because the object's axial motion changes the magnification.
Difficulty: Very hard
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