Light in air is incident at an angle of $45^{o}$ on the surface of glass plate having refractive index $1.526$ through what angle is light deviated upon refraction at top of surface.

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Solution

$∠ i = 45^{o}$
$^{a} u_{g} = 1.52$
Using Snell's law, $^{a} u_{g} = \frac{sin i}{sin r}$
$\Rightarrow sin r = \frac{sin i}{^{a} u_{g}}$
$\Rightarrow sin r = \frac{sin 45^{o}}{1.52} = \frac{0.707}{1.52}$
$\Rightarrow sin r = 0.46$
$\Rightarrow ∠ r = {sin}^{- 1} 0.46$

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Light in air is incident at an angle of 45o on the surface of glass plate having refractive index 1.526 through what angle is light deviated upon refraction at top of surface.

∠i=45oaug=1.52Using Snell's law, aug=sinisinr⇒sinr=siniaug⇒sinr=sin45o1.52=0.7071.52⇒sinr=0.46⇒∠r=sin−10.46

Light in air is incident at an angle of $45^{o}$ on the surface of glass plate having refractive index $1.526$ through what angle is light deviated upon refraction at top of surface.

$∠ i = 45^{o}$
$^{a} u_{g} = 1.52$
Using Snell's law, $^{a} u_{g} = \frac{sin i}{sin r}$
$\Rightarrow sin r = \frac{sin i}{^{a} u_{g}}$
$\Rightarrow sin r = \frac{sin 45^{o}}{1.52} = \frac{0.707}{1.52}$
$\Rightarrow sin r = 0.46$
$\Rightarrow ∠ r = {sin}^{- 1} 0.46$