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# A ray of light undergoes refraction and loses $0.25$ times its velocity. If 'i' is the angle of incidence and is very small, the angle of deviation for the ray of light is______

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Solution

## Step 1: Given dataThe angle of incidence $=i$${v}_{2}={v}_{1}-0.25{v}_{1}=0.75{v}_{1}$Step 2: Formula used${\mu }_{1}\mathrm{sin}i={\mu }_{2}\mathrm{sin}r$$\frac{{\mu }_{2}}{{\mu }_{1}}=\frac{{\nu }_{1}}{{v}_{2}}$$\delta =i-r$Step 3: SolutionLet $\delta$ be the angle of deviationBy Snell's law ${\mu }_{1}\mathrm{sin}i={\mu }_{2}\mathrm{sin}r......\left(1\right)$Also,$\frac{{\mu }_{2}}{{\mu }_{1}}=\frac{{\nu }_{1}}{{v}_{2}}......\left(2\right)\left[\because Refractiveindex=\frac{speedofthelightinvaccum}{speedoflightinthegivenmedium}\right]$From $\left(2\right)$$\frac{{\mu }_{2}}{{\mu }_{1}}=\frac{1}{0.75}$ Putting the above value in $\left(1\right)$$\mathrm{sin}r=0.75·\mathrm{sin}i$ $⇒r=0.75i$ $\left[\because if\theta issmall,\mathrm{sin}\left(\theta \right)\approx \theta \right]$So, the angle of deviation is, $\delta =i-r$$⇒\delta =i-0.75i\phantom{\rule{0ex}{0ex}}⇒\delta =0.25i$Hence, the angle of deviation is $0.25i$.

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