Question

A glass prism of angle $A=60^{{}^{\circ }}$ gives a minimum angle of deviation $\theta =30^{{}^{\circ }}$ 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. $\frac{100\pi }{180}\text{\%}$
• B. $\frac{5\pi }{180}\text{\%}$
• C. $\frac{50\pi }{180}\text{\%}$
• D. $\frac{5\pi }{18}\text{\%}$

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
$\frac{\text{d}\mu }{\text{dδ}}=\frac{cos\left(\frac{\delta +A}{2}\right)}{\text{2sin}\left(\frac{\text{A}}{2}\right)}$
$\Rightarrow \frac{\text{d}\mu }{\mu }=\frac{\text{cos }\left(\frac{\delta +A}{2}\right)}{\text{2sin }\left(\frac{\delta +A}{2}\right)}d\delta$
$\Rightarrow \frac{\text{d}\mu }{\mu }=\frac{1}{2}\text{cot}\left(\frac{\delta +A}{2}\right)\text{dδ}$
$\left(\frac{\delta +A}{2}\right)=45^{\circ }$
$\Rightarrow \frac{d\mu }{\mu }\times 100=\frac{1}{2}\cot \left(45^{\circ }\right)\frac{\pi }{180}\times 100$
Percentage error $=\frac{5\pi }{18}\%$