Page 184 - Physics_Form_2

Physics for Secondary Schools

Angle of minimum deviation of a prism The sum of internal angles is 180
The value of the angle of deviation D depends   
on the angle of incidence, i. As the angle of AEF  EAF  EFA  180
incidence increases, the angle of deviation 90 rA  90  180.....(6)
i
decreases to a minimum value called angle of 2
minimum deviation (Dm), and then starts to But we know that at a minimum
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increase again as shown in Figure 5.13.
r
angle of deviation, i = , then
2
Equation (6) can be written as,
r
r
90 
A
90 
180
Angle of deviation, D D m r = A 2r , this implies that,
45
A =+ =
rr
40
2
35
i=e



GFE 
180
GEF 
20 25 30 35 40 45 50 55 60 65 Now consider triangle GEF
EGF 
Angle of incidence, i
90  A
A 90
i−r +e−r +180°− D = 180° − D = 180°
90  i−r +e−r +180° 90r r 180   m m r  180
r
Figure 5.13: Variation of the angle of deviation
with the angle of incidence But, i = e
i−r +i−r +180°− D = 180° − D = 180°
A 90
90  i−r +e−r +180° 90r r 180   r  180
90  A
r
At the angle of minimum deviation, the angle of m m
incidence and the angle of emergence are equal. By simplifying you get,
That is, i = e. At the angle of minimum deviation, D =i−r +i−r
the refracted ray from the first surface travels m
through the prism perpendicular to the bisector of This becomes,
the apex angle A as shown in Figure 5.14. D = i−r +i−r = 2i−2r.
A m
This becomes,
D = 2i− A as r = A
N m 2
G M
D m Therefore,
Incident i i e
beam E r r Emergent beam i = AD+ m ............................(vii)
N' M' 2
B C Consider the first surface, by Snell’s
law
Figure 5.14: Deducing the angle of minimum  A  D  m
deviation sini sin   2  
η= 
sinr A
Consider triangle ABC in Figure 5.14. sin 2

178
Student’s Book Form Two

Physics Form 2 Final.indd 178 25/10/2025 10:28