A ray of light incident at an angle θ on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5° and the prism is made of a material of refractive index 1.5, the angle of incidence is

Question

A ray of light incident at an angle θ on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5° and the prism is made of a material of refractive index 1.5, the angle of incidence is (A) 7.5° (B) 5° (C) 15° (D) 2.5°

Tardigrade Admin · Accepted Answer

According to the question, ray emerges from other surface of prism normally,

∴

Angle of incidence at second face,

r^{'} = 0^{\circ}

Now,

r + r^{'} = A

\Rightarrow r = A - r^{'} = 5^{\circ} - 0^{\circ} = 5^{\circ}

Using snell's law

μ = \frac{s in i}{s in r}

or

s in i = μ s in r = 1.5 \times s in 5^{\circ}

= 0.131

\Rightarrow θ = i = s i n^{- 1} (0.131)

= 7. 5^{\circ}

Q. A ray of light incident at an angle $θ$ on a refracting face of a prism emerges from the other face normally. If the angle of the prism is $5^{\circ}$ and the prism is made of a material of refractive index $1.5$ , the angle of incidence is

$7. 5^{\circ}$

$5^{\circ}$

$1 5^{\circ}$

$2. 5^{\circ}$

Q. A ray of light incident at an angle θ on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5∘ and the prism is made of a material of refractive index 1.5, the angle of incidence is

7.5∘

5∘

15∘

2.5∘

According to the question, ray emerges from other surface of prism normally, ∴ Angle of incidence at second face, r′=0∘ Now, r+r′=A ⇒r=A−r′=5∘−0∘=5∘ Using snell's law μ=sinrsini​ or sini=μsinr=1.5×sin5∘ =0.131 ⇒θ=i=sin−1(0.131) =7.5∘

Q. A ray of light incident at an angle $θ$ on a refracting face of a prism emerges from the other face normally. If the angle of the prism is $5^{\circ}$ and the prism is made of a material of refractive index $1.5$ , the angle of incidence is

$7. 5^{\circ}$

$5^{\circ}$

$1 5^{\circ}$

$2. 5^{\circ}$