For a ray of light travelling from a denser, medium of refractive index n₁ to a rarer medium of refractive index n₂. Prove that n₂/ n₁= sin i_c where i_c is the critical angle of incidence for the media.

Solution

Given, a ray of light travelling from denser medium of refractive index n₁ to a rarer medium of refractive index n_2.Snell's law is given by,

$\frac{\sin i}{\sin r} = \frac{n_{2}}{n_{1}}$

Now, critical angle is the angle of incidence in denser medium for which angle of refraction in rarer medium is 90^o .
i.e., i = i_cwhen r = 90^o .

Thus, putting these values in snell's law , we have

$\frac{\sin i_{c}}{\sin 90^{o}} = \frac{n_{2}}{n_{1}}$

i.e., $\frac{n_{2}}{n_{1}} = \sin i_{c}$

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For a ray of light travelling from a denser, medium of refractive index n₁ to a rarer medium of refractive index n₂. Prove that n₂/ n₁= sin i_c where i_c is the critical angle of incidence for the media.

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Wave Optics

For a ray of light travelling from a denser, medium of refractive index n1 to a rarer medium of refractive index n2. Prove that n2 / n1= sin ic where ic is the critical angle of incidence for the media.

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Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

For a ray of light travelling from a denser, medium of refractive index n₁ to a rarer medium of refractive index n₂. Prove that n₂/ n₁= sin i_c where i_c is the critical angle of incidence for the media.