Question

Draw reflected rays for the following incident rays a. Incident ray assing through focus of a concave mirror b. Incident ray directed towards centre of curvature of a convex mirror

Answer 1

$\huge\underline\mathfrak\pink{Answer}$

✳️In the above ray diagram,

i = angleofincidence

r = angleofreflection

Answer 2

$\underline{\underline{\pink{\huge\sf Answer}}}$

The answers for both the questions have been attached.

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☯ $\underline{\textsf{Fermat's Principle of Least Time}}$

We all know that light travels in a straight line. So let's try to understand how is that happening. As per Fermat's Principle of Least Time,

✭ Between any two points say A & B light always takes the quickest path

✭ So we all know that the quickest path between any two points is a straight line

✭ And this gives us an idea why light travels in a straight line

✭ But we got to note that here it is mentioned the quickest path and not the shortest path, which means it is not mandatory that it has to be a straight line

We can even prove why Incident Angle is equal to the Reflected Angle (I = R) using this principle

✭ As we saw above light takes the shortest path and so it is mostly a straight line, which means that when an incident ray hits a surface and get's reflected in the shortest path

✭ Let's say there is two points A & B so if it such that there is a light ray coming from A and then it have to bounce somewhere and then reach point B and so after it is incident at some point say C which is the mid point of A & B then for it to reach B sill follow Fermat's Principle it has to reflect in the same Angle

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