By Team Aakash Byju's | 16th December 2022
n
Then the angles r, r’, e and the total deviation after two refractions are
Detailed Explanation
Applying Snell’s Law to the refraction of the light incident with an angle of 60° with an angle of refraction r is given as
n sin60 = N sinr
sin r =
. . r =
1
( )
√3
2
=
√3
sin r
sin 60 =
√3
2
. .
.
( )
1
√3
.
( )
√3
2
=
1
2
30
o
1
2
( )
sin 30 =
. .
.
n
o
o
o
Next, apply Snell’s Law at the second surface with the angle of incidence r’ and angle of emergence e given as:
n sin e = N sin r
r = r =
1
=
√3
sin r
30
o
. .
.
’
sin e
’
OA = OB, by similarity
’
n
sin e =√3 sin 30
. .
.
o
sin e =√3
( )
√3
2
1
2
=
. .
.
e = 60
o
sin 60 =
√3
2
( )
So far we calculated that, r = 30°, r’ = 30° and e = 60°. Now, let us find the total deviation.
o
n
Firstly, find the deviation at both surfaces.
Deviation at the first surface = i - r = 60 − 30 = 30°
Deviation at the second surface = i - r = 60 − 30 = 30°
o
o
o
o
n
Total deviation = deviation at first surface + deviation at second surface
δ =30° + 30° = 60°
total
n