Q. 22

# <span lang="EN-US

(i) Let’s assume the given postulate is true, then if A and B be two parallel rays entering to such a medium from air the refracted diagram will be FIG (1)

Here incidence angle is θi and refracted angle is θr . AB shows the incident wave front and ED shows the refracted wave front, then all the points on ED will have the same phase.

We know all the points with the same optical path length must have the same phase.

FIG (1)

Given n2=

n2AE=BC+ n2CD

If BC>0,

CD should be greater than 0, which is true according postulate diagram, which means rays of light enter such a medium from air (refractive index = 1) at an angle in 2nd quadrant, then the refracted beam is in the 3rd quadrant.

Let us consider the normal case where light entering 2nd quadrant refracting in 4rth quadrant as in FIG (2), then

n2AE=BC+ n2CD

,

which means if AE>CD, BC <0 which is not possible

the postulate is true

FIG (2)

(ii) From FIG (1)

--Hence Snell’s law is proved

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