KCET · Physics · Ray Optics
A ray of light passes through an equilateral glass prism in such a manner that the angle of incidence is equal to the angle of emergence and each of these angles is equal to \(\frac{3}{4}\) of the angle of the prism. The angle of deviation is
- A \(39^{\circ}\)
- B \(20^{\circ}\)
- C \(30^{\circ}\)
- D \(45^{\circ}\)
Answer & Solution
Correct Answer
(C) \(30^{\circ}\)
Step-by-step Solution
Detailed explanation
For an equilateral glass prism,
Angle of prism, \(A=60^{\circ}\)
If \(i\) and \(c\) be the angle of incidence and angle of cmergence respectively, then
\(i=\frac{3}{4} A=\frac{3}{4} \times 60^{\circ}=45^{\circ} \text { and } e=i=45^{\circ}\)
For a prism, we know that
\(\begin{aligned}A+\delta & =i+c \Rightarrow \delta=i+e-A \\\delta & =45^{\circ}+45^{\circ}-60^{\circ}=90^{\circ}-60^{\circ}=30^{\circ}\end{aligned}\)
Angle of prism, \(A=60^{\circ}\)
If \(i\) and \(c\) be the angle of incidence and angle of cmergence respectively, then
\(i=\frac{3}{4} A=\frac{3}{4} \times 60^{\circ}=45^{\circ} \text { and } e=i=45^{\circ}\)
For a prism, we know that
\(\begin{aligned}A+\delta & =i+c \Rightarrow \delta=i+e-A \\\delta & =45^{\circ}+45^{\circ}-60^{\circ}=90^{\circ}-60^{\circ}=30^{\circ}\end{aligned}\)
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