refraction of light

Question

Two rays emanating from a point source of light S, are perpendicular to each other. They pass from air to liquid. The first beam bends at an angle of 30˚, the second beam at an angle of 45˚. Find the liquid refractive index. The air refractive index is 1.

Answer 1

Diagram:

Given : r₁=30° and r₂=45° ,μ₁=1

To find : μ₂

Formula : Snell's law →μ₁sini=μ₂sinr 

Solution : As ∆SAB is right angled triangle,

\(\therefore\)  i₁+i₂= 90°

According to snell's law ,

For refraction at A,

  sini₁=μ₂sin30°

\(\therefore \) sini₁=μ₂(1/2)        .......(1)

For refraction at B ,

  sini₂=μ₂sin45°

\(\therefore\) sini₂=μ₂(1/√2)

\(\therefore\) sin(90°-i₁)=cosi₁=μ₂(1/√2) ...(2)

Square and add (1) and (2)

\(\therefore\) sin²i₁+cos²i₂=1=(μ₂)²(3/4)

\(\implies \,\mu _\textbf2=\frac{\textbf2}{\sqrt{\textbf3}}\) 

Refractive index of liquid is \(\frac{\textbf2}{\sqrt{\textbf3}}\)

Comments

Thank you very much!