NEET · Physics · STD 12 - 9. Ray optics and optical instruments
A ray of monochromatic light is passing through an equilateral prism (ABC) as shown in the figure. The refracted ray (QR) is parallel to its base (BC) and the angle of incidence (i) is \(50^{\circ}\). Then the angle of deviation \((\delta)\) is :
- A \(45^{\circ}\)
- B \(35^{\circ}\)
- C \(40^{\circ}\)
- D \(55^{\circ}\)
Answer & Solution
Correct Answer
(C) \(40^{\circ}\)
Step-by-step Solution
Detailed explanation
(C) \(40^{\circ}\)
For an equilateral prism, the angle of the prism is \(A=60^{\circ}\).
It is given that the refracted ray inside the prism is parallel to its base. This implies that the prism is in the position of minimum deviation.
In the position of minimum deviation, the angle of emergence \(e\) is equal to the angle of incidence \(i\).
Given \(i=50^{\circ}\), we have \(e =50^{\circ}\).
The angle of deviation \(\delta\) is given by the relation:
\(\delta=i+e-A\)
Substituting the values, we get:
\(\delta=50^{\circ}+50^{\circ}-60^{\circ}\)
\(\delta=100^{\circ}-60^{\circ}=40^{\circ}\)
For an equilateral prism, the angle of the prism is \(A=60^{\circ}\).
It is given that the refracted ray inside the prism is parallel to its base. This implies that the prism is in the position of minimum deviation.
In the position of minimum deviation, the angle of emergence \(e\) is equal to the angle of incidence \(i\).
Given \(i=50^{\circ}\), we have \(e =50^{\circ}\).
The angle of deviation \(\delta\) is given by the relation:
\(\delta=i+e-A\)
Substituting the values, we get:
\(\delta=50^{\circ}+50^{\circ}-60^{\circ}\)
\(\delta=100^{\circ}-60^{\circ}=40^{\circ}\)
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