The angle of a prism is 60 o and its refractive index is 2 . The angle of minimum deviation suffered by a ray of light in passing through it is

Refractive index, μ = sin ⁡ A + δ m 2 sin ⁡ A 2 ⇒ 2 = sin ⁡ 60 o + δ m 2 sin ⁡ 60 o 2 ⇒ 2 × s i n 30 o = sin ⁡ 60 o + δ m 2 ⇒ sin ⁡ 45 o = sin ⁡ 60 o + δ m 2 ⇒ δ m = 30 o

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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsTIR/Prism and DispersionEasy2 minPYQ_2020

PhysicsEasysingle choice

The angle of a prism is $60^{o}$ and its refractive index is $\sqrt{2} .$ The angle of minimum deviation suffered by a ray of light in passing through it is

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Solution:

Refractive index, $μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})}$

$\Rightarrow \sqrt{2} = \frac{\sin (\frac{60^{o} + δ_{m}}{2})}{\sin (\frac{60^{o}}{2})}$

$\Rightarrow \sqrt{2} \times s i n 30^{o} = \sin (\frac{60^{o} + δ_{m}}{2})$

$\Rightarrow \sin 45^{o} = \sin (\frac{60^{o} + δ_{m}}{2})$

$\Rightarrow δ_{m} = 30^{o}$

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

⏱ 2mℹ️ Source: PYQ_2020

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