A ray of light on a prism having one silvered surface, at an incident angle of $45^{\circ}$ as shown in figure. After refraction and reflection it retraces the path, then the refractive index of prism material is (prism angle is $30^{\circ}$ )

$\sqrt{2}$
2
$\frac{1}{\sqrt{2}}$
$\frac{1}{2}$

Question

A ray of light on a prism having one silvered surface, at an incident angle of $45^{\circ}$ as shown in figure. After refraction and reflection it retraces the path, then the refractive index of prism material is (prism angle is $30^{\circ}$ )

$\sqrt{2}$
2
$\frac{1}{\sqrt{2}}$
$\frac{1}{2}$

Toppr Admin · Accepted Answer

N1-1-n2-x3B8-1-450-x3B8-2-300 -by using geometry-we known1sin-x3B8-1-n2sin-x3B8-2n2-sin-x3B8-1sin-x3B8-2-sin450sin300n2-1-414-x221A-2hence- the-xA0-refractive index of prism material is -x221A-2

A ray of light on a prism having one silvered surface, at an incident angle of 45∘ as shown in figure. After refraction and reflection it retraces the path, then the refractive index of prism material is (prism angle is 30∘)√221√212

n1=1;n2=?;θ1=450;θ2=300 (by using geometry)we known1sinθ1=n2sinθ2n2=sinθ1sinθ2=sin450sin300n2=1.414=√2hence, the refractive index of prism material is √2

A ray of light on a prism having one silvered surface, at an incident angle of $45^{\circ}$ as shown in figure. After refraction and reflection it retraces the path, then the refractive index of prism material is (prism angle is $30^{\circ}$ )

$\sqrt{2}$
2
$\frac{1}{\sqrt{2}}$
$\frac{1}{2}$

$n_{1} = 1; n_{2} = ?; θ_{1} = 45^{0}; θ_{2} = 30^{0}$ (by using geometry)
we know
$n_{1} s i n θ_{1} = n_{2} s i n θ_{2}$
$n_{2} = \frac{s i n θ_{1}}{s i n θ_{2}} = \frac{s i n 45^{0}}{s i n 30^{0}}$
$n_{2} = 1.414 = \sqrt{2}$
hence, the refractive index of prism material is $\sqrt{2}$