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Angle of deviation $(\delta)$ by a prism (refractive index $=\mu$ and supposing the angle of prism $\mathrm{A}$ to be small) can be given by
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Verified Answer
The correct answer is:
$\delta=(\mu-1) \mathrm{A}$
Correct option is 1. $\delta=(\mu-1) A$
The refractive index of a prism is given by,
$\mu=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
For a very thin prism, we can assume that $\sin \theta \approx \theta$
$\begin{aligned} & \therefore \mu=\frac{\frac{A+\delta}{2}}{\frac{A}{2}} \\ & \text { or, } \mu=\frac{A+\delta}{A}\end{aligned}$
or, $\delta=\mu A-A$
or, $\delta=(\mu-1) A$ is the required angle of deviation.
The refractive index of a prism is given by,
$\mu=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
For a very thin prism, we can assume that $\sin \theta \approx \theta$
$\begin{aligned} & \therefore \mu=\frac{\frac{A+\delta}{2}}{\frac{A}{2}} \\ & \text { or, } \mu=\frac{A+\delta}{A}\end{aligned}$
or, $\delta=\mu A-A$
or, $\delta=(\mu-1) A$ is the required angle of deviation.
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