Question

A glass prism of angle $A=60^{^\circ }$ gives a minimum angle of deviation $\theta = 3 0^{^\circ }$ with the maximum error of $1^{^\circ }$ when a beam of parallel light passed through the prism during an experiment. The permissible error in the measurement of refractive index $\mu $ of the material of the prism is :

• A. $\frac{1 0 0 \pi }{1 8 0} \text{\%}$
• B. $\frac{5 \pi }{1 8 0} \text{\%}$
• C. $\frac{5 0 \pi }{1 8 0} \text{\%}$
• D. $\frac{5 \pi }{1 8} \text{\%}$

Answer 1

Answer: (D)

$\mu=\frac{\sin \left(\frac{\delta+ A }{2}\right)}{\left(\sin \left(\frac{ A }{2}\right)\right)},$ where $\delta$ is the minimum deviation , where $\delta$ is the minimum deviation
$\frac{\text{d} \mu }{\text{dδ}}=\frac{cos \left(\frac{\delta + A}{2}\right)}{\text{2sin} \left(\frac{\text{A}}{2}\right)}$
$\Rightarrow \frac{\text{d} \mu }{\mu }=\frac{\text{cos } \left(\frac{\delta + A}{2}\right)}{\text{2sin } \left(\frac{\delta + A}{2}\right)}d\delta$
$\Rightarrow \frac{\text{d} \mu }{\mu }=\frac{1}{2}\text{cot}\left(\frac{\delta + A}{2}\right)\text{dδ}$
$\left(\frac{\delta + A}{2}\right)=45^\circ $
$\Rightarrow \frac{\text{d} \mu }{\mu }\times 100=\frac{1}{2}\text{cot}\left(45 ^\circ \right)\frac{\pi }{180}\times 100$
Percentage error $= \frac{5 \pi }{1 8\%}$