Physics Paper 2, Nov/Dec. 2012  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
General Comments

Question 13

 Question 13
(a)   Define:
        (i)   principal focus for a diverging lens;
       (ii)   principal axis.
(b)    State a situation under which the speed of light can change.
(c)      The separation between the lenses of an astronomical telescope in normal
        adjustment is 35.0 cm and the angular magnification is 8

    •  Calculate the focal lengths of the lenses.
    • The lenses separation are adjusted to produce the final image 30 cm from the eye piece lens.  Calculate the new separation of the lenses
    • Sketch a ray diagram to illustrate the image formed in 13(c)(ii)above.

(a)    Most respondent candidates were not able to define correctly principal focus for a
        diverging lens and principal axis.

(b)    Relatively few candidates succeeded in stating the required situation under which
        the speed of light can change.  Many of them failed to realize that the speed
        changes as light travels from one medium to another of different optical densities.

(c)    Most candidates messed up with their responses on this numerical on astronomical telescope in normal adjustment.  This state of affairs could readily be ascribed to the effect of rote-learning as candidates reactions showed lack of proper understanding of the concept. The sketching of the ray diagram was a mirage.

The expected answers are:

(a)(i)    This is a point on the principal axis to which all rays parallel and close to the
           principal axis appear to diverge after refraction / passing through the lens.

    (ii)    This is a line joining the centre of curvature (principal focus) of the lens to
            the optical centre.   
(b)        When the light travels from one medium to another of different refractive
           index/optical density.
 (c)       Angular magnification=      fo                                                                    
                                    fo + fe              =          35
                                    8fe + fe                            =          35
                                    9fe                    =          35
                                    fe                     =          3.89 cm                                                         
                                    fo                     =          35 – 3.89
                                                            =          31.11 cm                                                       
                                    1                      =          1       +    1                                                     
                                    fe                                 u               v
                                    1                      =          1       +        1                                                 
                                  3.89                              u                -30

                                    1                      =          1       +   1       =     30 + 3.89                       
                                    u                                3.89         30               116.7

u                      =          116.7                                     

                                                            =          3.44cm                                                                      

New separation of lenses = 3.44 + 31.11 = 34.55cm                            

Parallel rays incident on objective lens
Correct position of real image formed by objective lens                                      
Virtual image formed close to the objective lens                       




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