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AIIMS 201827 May
Exam
AIIMS
Year
2018
Date
27 May
Subject
Physics
Chapter
Ray Optics

AIIMS 2018 · 27 May · Physics · Ray Optics

In the figure, a ray of light is perpendicular to the face A B of a...

In the figure, a ray of light is perpendicular to the face ABA B of a glass prism (μ=1.52)(\mu=1.52). In the figure, a ray of light is perpendicular to the face A B of a glass prism (μ=1.52) . Find the value of θ so that
Find the value of θ\theta so that the ray is totally reflected at face ACA C, if the prism is immersed in water (μ=1.33)(\mu=1.33). Given sin⁡61∘=0.875\sin 61^{\circ}=0.875

Options

  1. A45∘45^{\circ}
  2. B30∘30^{\circ}
  3. C15∘15^{\circ}
  4. DNone of these

Correct answer

B. 30∘30^{\circ}

Step-by-step solution

Given, μg=1.52,μw=1.33,θ=\mu_g=1.52, \mu_w=1.33, \theta= ? Apply Snell's law at point DD, In the figure, a ray of light is perpendicular to the face A B of a glass prism (μ=1.52) . Find the value of θ so that
μgsin⁡ic=μwsin⁡90∘⇒sin⁡ic=μwμg=1.331.52\begin{aligned} & \mu_g \sin i_c=\mu_w \sin 90^{\circ} \\ & \Rightarrow \sin i_c=\frac{\mu_w}{\mu_g}=\frac{1.33}{1.52}\end{aligned}\Rightarrow \quad i_c=\sin ^{-1}(1.331.52)=61∘\left(\frac{1.33}{1.52}\right)=61^{\circ} From figure, $\theta+i_c=90^{\circ} \Rightarrow \quad \theta=90^{\circ}-61^{\circ}=29^{\circ} \approx 30^{\circ}
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