The minimum deviation produced by a hollow prism filled with a certain liquid is found to be 30°. The light ray is also found to be refracted at angle of 30°. The refractive index of the liquid is

Question

The minimum deviation produced by a hollow prism filled with a certain liquid is found to be 30°. The light ray is also found to be refracted at angle of 30°. The refractive index of the liquid is (A) √2 (B) √3 (C) √(3/2) (D) (3/2)

Tardigrade Admin · Accepted Answer

For minimum angle of deviation for a prism A=2 r, ∴ A=60° Now μ=( sin (60+30/2)/ sin (60)2) =( sin 45°/ sin 30°)=(1/√2) × (2/1)=√2

Q. The minimum deviation produced by a hollow prism filled with a certain liquid is found to be $3 0^{\circ}$ . The light ray is also found to be refracted at angle of $3 0^{\circ}$ . The refractive index of the liquid is

$2$

$3$

$\frac{3}{2}$

$\frac{3}{2}$

For minimum angle of deviation for a prism
$A = 2 r, ∴ A = 6 0^{\circ}$
Now $μ = \frac{s i n \frac{60 + 30}{2}}{s i n \frac{60}{2}}$
$= \frac{s i n 4 5 ^{\circ}}{s i n 3 0 ^{\circ}} = \frac{1}{2} \times \frac{2}{1} = 2$

Q. The minimum deviation produced by a hollow prism filled with a certain liquid is found to be 30∘. The light ray is also found to be refracted at angle of 30∘. The refractive index of the liquid is

2​

3​

23​​

23​

For minimum angle of deviation for a prism A=2r,∴A=60∘ Now μ=sin260​sin260+30​​ =sin30∘sin45∘​=2​1​×12​=2​

Q. The minimum deviation produced by a hollow prism filled with a certain liquid is found to be $3 0^{\circ}$ . The light ray is also found to be refracted at angle of $3 0^{\circ}$ . The refractive index of the liquid is

$2$

$3$

$\frac{3}{2}$

$\frac{3}{2}$

For minimum angle of deviation for a prism
$A = 2 r, ∴ A = 6 0^{\circ}$
Now $μ = \frac{s i n \frac{60 + 30}{2}}{s i n \frac{60}{2}}$
$= \frac{s i n 4 5 ^{\circ}}{s i n 3 0 ^{\circ}} = \frac{1}{2} \times \frac{2}{1} = 2$