A prism of angle A = 6 0 ∘ has minimum deviation θ = 3 0 ∘ with maximum error 1 ∘ . Treat A as exact. Find the permissible percentage error in its refractive index, to two decimal places.

μ = sin ( A /2 ) sin [( A + θ ) /2 ] . μ Δ μ = 2 1 cot ( 2 A + θ ) Δ θ = 2 1 180 π . Percentage error = 18 5 π % = 0.87266 … % → 0.87% .

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Physics - Error Question with Solution | TestHub

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A prism of angle $A = 6 0^{\circ}$ has minimum deviation $θ = 3 0^{\circ}$ with maximum error $1^{\circ}$ . Treat A as exact. Find the permissible percentage error in its refractive index, to two decimal places.

Answer:

0.87

Solution:

$μ = \frac{sin [( A + θ ) /2 ]}{sin ( A /2 )}$ .

$\frac{Δ μ}{μ} = \frac{1}{2} cot (\frac{A + θ}{2}) Δ θ = \frac{1}{2} \frac{π}{180}$ .

Percentage error $= \frac{5 π}{18} % = 0.87266 \dots % \to 0.87%$ .

Stream:JEESubject:PhysicsTopic:ErrorSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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