A ray of light travels from an optically dens | vedclass

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(C) When a light ray travels from a denser to a rarer medium,it undergoes refraction. The angle of deviation $\delta$ for refraction is given by $\del

Vedclass · Accepted Answer

$\pi-2C$

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(C) When a light ray travels from a denser to a rarer medium,it undergoes refraction. The angle of deviation $\delta$ for refraction is given by $\delta = |r - i|$,where $i$ is the angle of incidence and $r$ is the angle of refraction. Since $r > i$,$\delta = r - i$. As $i$ increases,$r$ also increases. The maximum value of $i$ is the critical angle $C$,at which point $r = 90^\circ$ or $\frac{\pi}{2}$ radians. Thus,the maximum deviation for refraction is $\delta_{\max} = \frac{\pi}{2} - C$. However,if the angle of incidence $i$ exceeds $C$,the ray undergoes total internal reflection. In this case,the angle of reflection is equal to the angle of incidence $(r = i)$. The deviation for reflection is $\delta = \pi - 2i$. To find the maximum deviation for reflection,we consider the range $C < i < \frac{\pi}{2}$. As $i$ approaches $C$ from the denser side,$\delta$ approaches $\pi - 2C$. This is the maximum possible deviation for the ray in this scenario.

$A$ ray of light travels from an optically denser to a rarer medium. The critical angle for the two media is $C$. The maximum possible deviation of the ray will be

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