Question

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a light incident at an interface at an angle of 60 degree is refracted into another medium at an angle of 30 degree. if its speed in the 1st medium is c, then its speed in second medium is what? explain how.
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Answer 1

▶ Solution :

✴ Given :

▪ Angle of incident = 60°

▪ Angle of refraction = 30°

▪ Velocity of light in first medium = C

✴ To Find :

▪ Velocity of light in second medium.

✴ Concept :

✏ This question is completely based on concept of snell's law.

✏ Refractive index of a medium is defined as the ratio of velocity of light in vacuum to the velocity of light in medium.

✴ Calculation :

$\implies\bf\:n_1\sin{i}=n_2\sin{r}\\ \\ \implies\sf\:\dfrac{c}{c}\times \sin60\degree=\dfrac{c}{v}\times \sin30\degree\\ \\ \implies\sf\:\dfrac{\sqrt{3}}{2}=\dfrac{c}{2v}\\ \\ \implies\sf\:v=\dfrac{c}{\sqrt{3}}\\ \\ \implies\sf\:v=\dfrac{3\times 10^8}{\sqrt{3}}\\ \\ \implies\boxed{\bf{\purple{v=1.73\times 10^8\:mps}}}$

Answer 2

Given,

➡️ Angle of incident = 60°

➡️Angle of refraction = 30°

➡️ Velocity of light (First medium)=C

To find:-

Velocity of the light in second medium.

Now,

n1 sin i = n2 sin r

$\frac{c}{c} \times sin60 = \frac{c}{c } \times sin30$

$\frac{ \sqrt{3} }{2} = \frac{c}{2v}$

$v = \frac{c}{ \sqrt{3} }$

$v = \frac{3 \times {10}^{8} }{ \sqrt{3} }$

so,

The answer is...!!

$v = 1.73 \times {10}^{8} mps$