The critical angle of light going from medium A into medium B is $θ$ . If the speed of light in medium A is $v$ , then the speed of light in medium B is :

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Answer 1

For total internal reflection
$n_{A} s i n θ = n_{B}$

when $n_{A} =$ refractive index of medium A

$n_{B} =$ refractive index of medium B

$\frac{n _{A}}{n _{B}} = \frac{1}{s i n θ}$

by snell's law

$\frac{n _{A}}{n _{B}} = \frac{v _{B} ( v e l o c i t y i n m e d i u m B )}{v _{A} ( v e l o c i t y i n m e d i u m A )}$

$\frac{v _{B}}{v _{A}} = \frac{1}{s i n θ}$

$v_{B} = \frac{v _{A}}{s i n θ}$

$= \frac{v}{s i n θ}$

Answer 2

For total internal reflection

n_{A} s i n θ = n_{B}

when

n_{A} =

refractive index of medium A

n_{B} =

refractive index of medium B

\frac{n _{A}}{n _{B}} = \frac{1}{s i n θ}

by snell's law

\frac{n _{A}}{n _{B}} = \frac{v _{B} ( v e l o c i t y i n m e d i u m B )}{v _{A} ( v e l o c i t y i n m e d i u m A )}

\frac{v _{B}}{v _{A}} = \frac{1}{s i n θ}

v_{B} = \frac{v _{A}}{s i n θ}

= \frac{v}{s i n θ}

The critical angle of light going from medium A into medium B is $θ$ . If the speed of light in medium A is $v$ , then the speed of light in medium B is :

$\frac{v}{sin θ}$

$v sin θ$

$\frac{v}{tan θ}$

$v tan θ$

The speed of light in two media I and II are $2.2 \times 1 0^{8} m / s$ and $2.4 \times 10^{8} m / s$ respectively. The critical angle for light refracting from I to II medium will be

The critical angle for light going from a medium in which wavelength is $4000 A^{\circ}$ to medium in which its wavelength is $6000 A^{\circ}$ is :

The critical angle of a medium B with respect another medium A is 45 $^{\circ}$ and the critical angle of a medium C with respect the medium B is 30 $^{\circ}$ . The critical angle of medium C with respect to A is :

Calculate the speed of light in a medium, whose critical angle is 45 $^{o}$ .

If the critical angle of the medium is $3 0^{\circ}$ , the velocity of light in that medium is :
(velocity of light in air $3 \times 1 0^{8}$ m/s)

Revise with Concepts

Total Internal Reflection and Critical angle

Optical Fibres

Critical Angle - Formula

Applications of Total Internal Reflection

Prism binoculars

The critical angle of light going from medium A into medium B is θ . If the speed of light in medium A is v, then the speed of light in medium B is :

sinθv​

vsinθ

tanθv​

vtanθ

For total internal reflection nA​sinθ=nB​ when nA​= refractive index of medium A nB​= refractive index of medium B nB​nA​​=sinθ1​ by snell's law nB​nA​​=vA​(velocity in medium A)vB​(velocity in medium B)​ vA​vB​​=sinθ1​ vB​=sinθvA​​ =sinθv​

The speed of light in two media I and II are 2.2×108m/s and 2.4×108m/s respectively. The critical angle for light refracting from I to II medium will be

The critical angle for light going from a medium in which wavelength is 4000A∘ to medium in which its wavelength is 6000A∘ is :

The critical angle of a medium B with respect another medium A is 45∘ and the critical angle of a medium C with respect the medium B is 30∘. The critical angle of medium C with respect to A is :

Calculate the speed of light in a medium, whose critical angle is 45o.

If the critical angle of the medium is 30∘, the velocity of light in that medium is : (velocity of light in air 3×108 m/s)

Revise with Concepts

Total Internal Reflection and Critical angle

Optical Fibres

Critical Angle - Formula

Applications of Total Internal Reflection

Prism binoculars

The critical angle of light going from medium A into medium B is $θ$ . If the speed of light in medium A is $v$ , then the speed of light in medium B is :

$\frac{v}{sin θ}$

$v sin θ$

$\frac{v}{tan θ}$

$v tan θ$

The speed of light in two media I and II are $2.2 \times 1 0^{8} m / s$ and $2.4 \times 10^{8} m / s$ respectively. The critical angle for light refracting from I to II medium will be

The critical angle for light going from a medium in which wavelength is $4000 A^{\circ}$ to medium in which its wavelength is $6000 A^{\circ}$ is :

The critical angle of a medium B with respect another medium A is 45 $^{\circ}$ and the critical angle of a medium C with respect the medium B is 30 $^{\circ}$ . The critical angle of medium C with respect to A is :

Calculate the speed of light in a medium, whose critical angle is 45 $^{o}$ .

If the critical angle of the medium is $3 0^{\circ}$ , the velocity of light in that medium is :
(velocity of light in air $3 \times 1 0^{8}$ m/s)