The correct option is
A A−(p,s),B−q,C−(r,s),D−qCase
A :-
Calculating critical angle for the two mediums :-
n1sinic=n2sin90∘
2×sinic=1×1
sinic=12
⇒ic=30∘
If angle of incidence is greater than
30∘ light ray will get reflected back from interface.
In first case
i>ic
⇒ Light will get refelcted back.
Angle of deviation :-
δ1=180∘−2×37∘
=180∘−74∘
=106∘
In case
B :-
TIR will not take place.
Hence, the angle of deviation will be less than
90∘
In case
C:-
i>ic
∴ Angle of deviation is given by:
δ3=180∘−2×45∘
=180∘−90∘=90∘
In case
D :-
Applying snell`s law for first referaction :-
n1sini=n2sinr
1×sin(90∘)=2×sinθ
(Grazing incidence)
⇒sinθ=12
or,
θ=30∘
Now, applying snell`s law at the second interface :-
2×sin30∘=√2sini′
sini′=1√2
or,
i′=45∘= angle of emergence
∴ Angle of deviation is given by :-
δ=180∘−i−e
=180∘−90∘−45∘=45∘
Here
e=i′)
In case
(A) and
(C) light is reflected back to the same medium.
Therefore, speed of ray will remain same in these two cases.
In the order two cases light goes to other medium after refraction hence, speed of ray will not remain same.
Hence, the correct answer is :-
A−(p,s),B−q,C−(r,s),D−q