A, B and C are the parallel sided transparent media of refractive index n1, n2 and n3 respectively. They are arranged as shown in the figure. A ray is incident at an angle θ on the surface of separation of A and B which is as shown in the figure. After the refraction into the medium B, the ray grazes the surface of separation of the media B and C. Then, Sin θ = .........

Question

A, B and C are the parallel sided transparent media of refractive index n1, n2 and n3 respectively. They are arranged as shown in the figure. A ray is incident at an angle θ on the surface of separation of A and B which is as shown in the figure. After the refraction into the medium B, the ray grazes the surface of separation of the media B and C. Then, Sin θ = ......... (A)  (n3/n1) (B)  (n1/n2) (C)  (n2/n3) (D)  (n1/n3)

Tardigrade Admin · Accepted Answer

Applying Snell's law between the surfaces A and B

n_{1} sin i = n_{2} sin r_{1}

...(i)
Again applying Snell's law between surfaces

\underline{B}

and

C

n_{2} sin r_{1} = n_{3} sin r_{2}

...(ii)
From Eqs. (i) and (ii), we get

n_{1} sin i = n_{3} sin r_{2}

Here,

r_{2} = 9 0^{\circ}

∴ n_{1} sin i = n_{3}

\Rightarrow sin i = \frac{n _{3}}{n _{1}}

$\frac{n _{3}}{n _{1}}$

$\frac{n _{1}}{n _{2}}$

$\frac{n _{2}}{n _{3}}$

$\frac{n _{1}}{n _{3}}$

n1​n3​​

n2​n1​​

n3​n2​​

n3​n1​​

Applying Snell's law between the surfaces A and B n1​sini=n2​sinr1​...(i) Again applying Snell's law between surfaces B​ and C n2​sinr1​=n3​sinr2​...(ii) From Eqs. (i) and (ii), we get n1​sini=n3​sinr2​ Here, r2​=90∘ ∴n1​sini=n3​ ⇒sini=n1​n3​​

$\frac{n _{3}}{n _{1}}$

$\frac{n _{1}}{n _{2}}$

$\frac{n _{2}}{n _{3}}$

$\frac{n _{1}}{n _{3}}$