A ray of light travelling in water is inciden | vedclass

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(C) When a light ray travels from a denser medium (water) to a rarer medium (air) and the angle of incidence $	heta$ is less than the critical angle,

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a reflected ray and a refracted ray and the angle between them would be less than $180^o - 2	heta$

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(C) When a light ray travels from a denser medium (water) to a rarer medium (air) and the angle of incidence $	heta$ is less than the critical angle,both reflection and refraction occur.According to the law of reflection,the angle of reflection is equal to the angle of incidence,which is $	heta$.According to Snell's law,the angle of refraction $r$ is greater than the angle of incidence $	heta$ because the light is moving from a denser to a rarer medium $(r > 	heta)$.The angle between the reflected ray and the surface is $90^o - 	heta$.The angle between the refracted ray and the surface is $90^o - r$.The angle between the reflected ray and the refracted ray is the sum of the angle between the reflected ray and the surface,the surface itself $(180^o)$,and the angle between the refracted ray and the surface is not the correct way to visualize it. Instead,consider the straight line representing the interface.The angle between the reflected ray and the refracted ray is $180^o - (	ext{angle of reflection} + 	ext{angle of refraction}) = 180^o - (	heta + r)$.Since $r > 	heta$,it follows that $	heta + r > 2	heta$.Therefore,$180^o - (	heta + r) Thus,the angle between them is less than $180^o - 2	heta$.

$A$ ray of light travelling in water is incident on its surface open to air. The angle of incidence is $\theta$,which is less than the critical angle. Then there will be

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