Higher Homework Physics Exercise 15-Refraction of light
Resources & Mark scheme
Resources:
Questions: / Past papers1 / Q16 – 2000
2 / Q16-2001
3 / Q17- 2000
4 / Q16-2003
5 / Q14-2005
6 / Q16- 2005
7 / Q27-2000
8 / Q27-2001
9 / Q27 (b)- 2003
10 / Q 27-2002
11 / Q27- 2004
Mark Scheme:
Q1 – C
Q2 – D
Q3 – C
Q4 – B
Q5 – B
Q6 – A
Q7 :
c.i.
The critical angle is found by noting the incident angle at which
the reflected intensity reaches a maximum.
θ= 42º
c.ii.
nglass = 1/sinθ critical
nglass= 1/sin42º
nglass = 1.49
c.iii.
The intensity of ray T will decrease as angle θis increased upto 42º.
At angles equal to and above 42º the intensity of ray T will fall
to zero, as the incident ray will be totally internally reflected.
Q8:
a) dsinθ= nλ
d = 2.16x10-6m
n = 2
λ= 486x10-9m
θ= ?
sinθ= nλ/d
sinθ= 2x486x10-9/2.16x10-6
sinθ= 0.45
θ= 26.74o
b.i)
Angle i = 47o
Angle r = 27o
nglass = sin(i)/sin(r)
nglass= sin47o/sin27o
nglass= 0.731/0.454
nglass= 1.61 ...as required
b.ii)θcritical= sin-1(1/n)
θcritical= sin-1(1/1.61)
θcritical= sin-1(0.613)
θcritical= 38.4o
At point X the incident angle of 63o is greater than
the critical angle. This means that the light is totally
internally reflected at this boundary.
Q9:
Q10:
a)nglass= sinθair/sinθglass
nglass= sin20o/sin13o
nglass= 1.52
b) The critical angle is the angle,measured between the ray
and the normal, at which light striking the glass air
boundary will be totally internally reflected.
c)
θcritical= sin-1(1/n)
θcritical= sin-1(1/1.52)
θcritical= sin-1(0.658)
θcritical= 41.1º
d)
Q11:
END OF EXERCISE 15