First slide

Prism

Question

The refractive index of material of prism is given by $μ = \frac{\sin \frac{(A + D_{m})}{2}}{\sin \frac{A}{2}}, A & D_{m}$ have their usual meanings. A glass prism of $A = 60^{0}$ gives minimum angle of deviation at $30^{0}$ with maximum error of $1^{0}$ , when a beam of parallel light passed through prism during experiment. Then percentage error in the measurement of $μ$ of material of prism is

Moderate

A

0.87%
B

2.22%
C

3.11%
D

1.65%

Solution

$\begin{array}{l} \Rightarrow μ = \frac{S i n \frac{(A + δ)}{2}}{S i n \frac{A}{2}} = \frac{S i n \frac{(60 + 30)}{2}}{S i n \frac{60}{2}} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \sqrt{2} \\ After differentiating we get, \end{array}$
$d μ = \frac{1}{\sin (\frac{A}{2})} \cos (\frac{A + D_{m}}{2}) . (\frac{d δ}{2}) = \cos e c (\frac{A}{2}) \cos (\frac{A + D_{m}}{2}) . (\frac{d δ}{2}) = \cos e c (\frac{60}{2}) \cos (\frac{60 + 30}{2}) . (\frac{1 °}{2}) = \cos e c (30) \cos (45) . (\frac{1 °}{2}) \times \frac{π}{180} = \frac{π}{180 \sqrt{2}}$
$\frac{d μ}{μ} \times 100 = \frac{\frac{π}{180 \sqrt{2}}}{\sqrt{2}} \times 100 = \frac{π}{360} \times 100 = 0.872 %$

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