Higher Homework Physics Exercise 15-Refraction of light

Resources & Mark scheme

Resources:

Questions: / Past papers
1 / 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