A ray of light is incident on a medium of ref | vedclass

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Question

(A) The angle of deviation $\delta$ for refraction is given by $\delta = i - r$,where $i$ is the angle of incidence and $r$ is the angle of refraction

Vedclass · Accepted Answer

$\frac{1}{\mu} = \cos \delta - \frac{\sin \delta}{	an i}$

Vedclass · Answer

(A) The angle of deviation $\delta$ for refraction is given by $\delta = i - r$,where $i$ is the angle of incidence and $r$ is the angle of refraction.From Snell's Law,$\mu = \frac{\sin i}{\sin r}$,which implies $\sin r = \frac{\sin i}{\mu}$.Since $\delta = i - r$,we have $r = i - \delta$.Substituting $r$ in the expression for $\mu$:$\mu = \frac{\sin i}{\sin(i - \delta)}$Taking the reciprocal:$\frac{1}{\mu} = \frac{\sin(i - \delta)}{\sin i}$Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$\frac{1}{\mu} = \frac{\sin i \cos \delta - \cos i \sin \delta}{\sin i}$$\frac{1}{\mu} = \frac{\sin i \cos \delta}{\sin i} - \frac{\cos i \sin \delta}{\sin i}$$\frac{1}{\mu} = \cos \delta - \frac{\sin \delta}{	an i}$

$A$ ray of light is incident on a medium of refractive index $\mu$ at an angle of incidence $i$. On refraction in the medium,$\delta$ is the angle of deviation. Then:

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