Find the critical angle for a ray of light travelling from glass to air- - g -2- - a -1-A- cannot be determinedB- 45-C- 90-D- 30-

The correct option is B 45o Apply Snell's law : √(2) sin i_c = (1) sin 90^∘ i_c = sin^ 1 (1/√(2)) i_c = 45^∘ ...

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Standard X

Physics

Refraction at Planar Surface

Find the crit...

Question

Find the critical angle for a ray of light travelling from glass to air. [ $μ_{g}$ = $\sqrt{2}$ , $μ_{a}$ = 1]

90^o

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45^o

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30^o

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cannot be determined

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Solution

The correct option is B
45^o

Apply Snell's law :

$\sqrt{2}$ $sin$ $i_{c}$ = (1) sin $90^{\circ}$

$i_{c}$ = ${sin}^{- 1}$ ( $\frac{1}{\sqrt{2}}$ )

$i_{c}$ = $45^{\circ}$

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Similar questions

Find the critical angle for a ray of light travelling from glass to air?

[ $μ_{g}$ = $\sqrt{2}$ , $μ_{a}$ = 1 ]

Q. Find the angle of refraction for a ray of light travelling from glass to air. Given angle of incidence is

30^{\circ}

and refractive index of glass is

μ_{g} = \sqrt{2}

Q. Critical angle for a light ray travelling from denser to rarer medium is defined as the angle of incidence at which the angle of refraction is 90°.
If a ray of light travels from water

(μ_{w})

to air

(μ_{a})

. Find the critical angle?

The critical angle for a light travelling from denser to rarer medium is defined as the angle of incidence at which the light refracts at $90^{\circ}$ (angle of refraction). If a ray of light travels from water ( $μ_{w}$ ) to air ( $μ_{a}$ ), find the critical angle.

Q. Critical angle for a light ray travelling from denser to rarer medium is defined as the angle of incidence at which the angle of refraction is 90°.
If a ray of light travels from water

(μ_{w})

to air

(μ_{a})

. Find the critical angle?

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Find the critical angle for a ray of light travelling from glass to air. [μg = √2 , μa = 1]

The correct option is B 45o Apply Snell's law : √2 sin ic = (1) sin 90∘ ic = sin−1 (1√2) ic = 45∘

Find the critical angle for a ray of light travelling from glass to air. [ $μ_{g}$ = $\sqrt{2}$ , $μ_{a}$ = 1]

The correct option is B
45^o

Apply Snell's law :

$\sqrt{2}$ $sin$ $i_{c}$ = (1) sin $90^{\circ}$

$i_{c}$ = ${sin}^{- 1}$ ( $\frac{1}{\sqrt{2}}$ )

$i_{c}$ = $45^{\circ}$