For a prism of refracting angle A , the refractive index is cot ⁡ A 2 . Then the minimum angle of deviation is

cot ⁡ A 2 = sin ⁡ A + δ m 2 sin ⁡ A 2 cos ⁡ A 2 sin ⁡ A 2 = sin ⁡ A + δ m 2 sin ⁡ A 2 sin ⁡ 90 - A 2 = sin ⁡ A + δ m 2 Hence, δ m = 180 ° - 2 A

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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsTIR/Prism and DispersionEasy2 minPYQ_2020

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For a prism of refracting angle $A$ , the refractive index is $\cot \frac{A}{2}$ . Then the minimum angle of deviation is

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Answer:

Solution:

$\cot \frac{A}{2} = \frac{\sin (\frac{A + δ_{m}}{2})}{\frac{\sin A}{2}}$

$\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}} = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}}$

$\sin (90 - \frac{A}{2}) = \sin (\frac{A + δ_{m}}{2})$

Hence, $δ_{m} = 180 ° - 2 A$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Geometrical OpticsSubtopic:TIR/Prism and Dispersion

⏱ 2mℹ️ Source: PYQ_2020

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