Answers to Light

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Answers to Light 2007NYGH/Sec3IP/EOYP2Q5 1(a) C Fig. 5.1 (b) (i) IA IB (ii) B appears to run towards A/C along a straight line. 2. a (i) Critical angle = sin -1 (1/1/5) = 42 o [1] (ii) Angle of refraction = sin -1 (sin 40 o /1.5) = 25 o [1] (iii) Angle of emergence = 50 o [1] Mirror M A B Fig. 5.2 C ray from I B to A---- 1 mark ray from B to C-----1 mark dotted ray from I B to C----1 mark Mirror M Dance studio wall Dance studio wall Two image drawn correctly award 1 mark 1

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Transcript of Answers to Light

  • Answers to Light 2007NYGH/Sec3IP/EOYP2Q5 1(a)

    C Fig. 5.1 (b) (i)

    IA IB

    (ii) B appears to run towards A/C along a straight line. 2. a (i) Critical angle = sin-1(1/1/5) = 42o [1] (ii) Angle of refraction = sin-1 (sin 40o/1.5) = 25o [1] (iii) Angle of emergence = 50o [1]

    Mirror M

    A B

    Fig. 5.2

    C

    ray from IB to A---- 1 mark ray from B to C-----1 mark dotted ray from IB to C----1 mark

    Mirror M Dance studio wall Dance studio wall

    Two image drawn correctly award 1 mark

    1

  • (iv)

    b. The prism does not have a silver backing like mirror which may peel off with age. The prism allows more light energy (stronger intensity) to enter the room. Multiple-reflections will occur at glass surfaces of the mirror which will reduce the light intensity entering the room.

    c. The setup will allow light rays that strike the surface from angle of incidence from 0o (shown by

    the blue ray) to an angle of 90o(shown by the red ray) to enter the room.

    Fig. 10.2

    50

    A

    B C 45

    25o

    70o

    70o

    25o

    50o

    45

    A

    B C

  • 2007NYGH/Sec3IP/EOYP2Q6 3 The diagram in Fig. 6 shows a ray of light from a point object O striking a converging lens L. The lens forms an image at point I.

    b. An upright, virtual and magnified image is formed. 2007NYGH/Sec4IP/EOYP2Q4 4 n= 1/sin c

    1.6 = 1/sin c c= sin-1(1/1.6) =39 n = sin i /sin r 1.6 = sin i /sin 21 i = sin-1 [1.6x sin 21] = 35 Path of incident ray drawn correctly Path of refracted ray drawn with correct angle r

    B1 B1 B1 B1

    2007NYGH/Sec4IP/EOYP2Q5 5(a) f = 12 cm A1 (b) 2 types of rays drawn

    One ray through centre One ray parallel to the centre ray correct symbol of the lens, image, arrow heads, and principal axis

    B1 B1

    (c) 1/u + 1/v = 1/f 1/24 + 1/v = 1/12 v = 24 cm

    B1 B1

    2007NYGH/Sec4Olvl/PrelimP2Q4 6 (a) Critical angle is incidence angle in optically denser medium for which the angle of

    refraction is 90 / refracted ray emerges along the boundary of the 2 media. [1]

    O

    I

    X

    Fig. 6

    L

    60 35

    21 39

    Lens

    Image

    Focal plane

    Parallel rays from distant object

    F

  • (b) cw = 39 [1] R.I. = 1/sin cw [1] = 1.59 [1] (ignore s.f.)

    (c) No penalty if angles not labelled. If angles labelled correctly but drawn wrongly, accept it as correct. 1m for refraction at Q and T (correct angles of refraction determined) 1m for reflection at R ( i = r) 1m for total internal reflection at S (i=r)

    P

    Q

    28.8 48 48 67 67 23 38.4

    R

    S

    T

    50