
A light ray passes from a material of low refractive index to one of high refractive index. Which of the pairs of quantities listed below describes the light ray as it strikes and passes through the interface between the two materials? Match the descriptions below with the statement that best describes the situation.
A part of the light ray remains inside the low refractive index material:
(A) The angle of refraction is larger than the angle of incidence
(B) The angle of refraction is equal to the angle of incidence
(C) The angle of reflection is larger than the angle of incidence
(D) The angle of incidence equals angle of reflection
(E) The angle of reflection is smaller than angle of incidence
Answer
233.1k+ views
Hint Understand and visualize the concept of refraction of light. Apply snell’s law for the given case and derive at your answer.
Complete step by step solution
It is given that a light ray passes from a material of low refractive index to high refractive index. Therefore $n_1$<$n_2$.
According to snell’s law of refraction,
\[{n_1}\sin {\theta _i} = {n_2}\sin {\theta _r}\]
Where \[\sin {\theta _i}\] is the angle of incidence and \[\sin {\theta _r}\] is the angle of refraction.
When the light ray passes from high to low medium, i.e. $n_1$>$n_2$, the angle of refraction will always be greater than the angle of incidence.
But in our case, it is mentioned that light rays move from low to high medium, hence $n_1$<$n_2$. In this case, the angle of refraction is smaller than that of angle of incidence.
The refraction angle changes when the light ray changes medium, but the angle of reflection will always be the same as angle of incidence. There won’t be any effect on reflection due to change in medium, hence the angle of reflection will be always equal to angle of reflection at any medium, since both takes place in the same medium.
Hence, option (d) is the right answer for the given question.
Note
Snell’s law is the mathematical relation that mentions the relationship between the path taken by the light ray passing through a boundary of two different mediums. The mathematical equation relates the angle of incidence in the first medium and the deviation or refraction taken by the light ray when it hits the boundary of the second medium. Refraction always happens when there’s a change of medium.
Complete step by step solution
It is given that a light ray passes from a material of low refractive index to high refractive index. Therefore $n_1$<$n_2$.
According to snell’s law of refraction,
\[{n_1}\sin {\theta _i} = {n_2}\sin {\theta _r}\]
Where \[\sin {\theta _i}\] is the angle of incidence and \[\sin {\theta _r}\] is the angle of refraction.
When the light ray passes from high to low medium, i.e. $n_1$>$n_2$, the angle of refraction will always be greater than the angle of incidence.
But in our case, it is mentioned that light rays move from low to high medium, hence $n_1$<$n_2$. In this case, the angle of refraction is smaller than that of angle of incidence.
The refraction angle changes when the light ray changes medium, but the angle of reflection will always be the same as angle of incidence. There won’t be any effect on reflection due to change in medium, hence the angle of reflection will be always equal to angle of reflection at any medium, since both takes place in the same medium.
Hence, option (d) is the right answer for the given question.
Note
Snell’s law is the mathematical relation that mentions the relationship between the path taken by the light ray passing through a boundary of two different mediums. The mathematical equation relates the angle of incidence in the first medium and the deviation or refraction taken by the light ray when it hits the boundary of the second medium. Refraction always happens when there’s a change of medium.
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