A ray of light passes from a medium $A$ having refractive index $1.6$ to the medium $B$ having refractive index $1.5$ . The value of critical angle of medium $A$ is _____

{sin}^{- 1} \sqrt{\frac{16}{15}}

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{sin}^{- 1} (\frac{16}{15})

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{sin}^{- 1} (\frac{1}{2})

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{sin}^{- 1} (\frac{15}{16})

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Solution

The correct option is D ${sin}^{- 1} (\frac{15}{16})$
As the law of refraction we have,
$\frac{sin i_{c}}{sin 90^{\circ}} = \frac{1.5}{1.6}$
$∴ sin i_{c} = \frac{15}{16}$
$\Rightarrow i_{c} =$ Critical angle of medium $A$
$= {sin}^{- 1} (\frac{15}{16})$ .

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A ray of light passes from a medium A having refractive index 1.6 to the medium B having refractive index 1.5. The value of critical angle of medium A is _____

The correct option is D sin−1(1516)As the law of refraction we have,sinicsin90∘=1.51.6∴sinic=1516⇒ic= Critical angle of medium A=sin−1(1516).

A ray of light passes from a medium $A$ having refractive index $1.6$ to the medium $B$ having refractive index $1.5$ . The value of critical angle of medium $A$ is _____

The correct option is D ${sin}^{- 1} (\frac{15}{16})$
As the law of refraction we have,
$\frac{sin i_{c}}{sin 90^{\circ}} = \frac{1.5}{1.6}$
$∴ sin i_{c} = \frac{15}{16}$
$\Rightarrow i_{c} =$ Critical angle of medium $A$
$= {sin}^{- 1} (\frac{15}{16})$ .