Two convex lenses of power 2 D and 5 D are separated by a distance 3 1 m . The power of optical system formed is

Given: P 1 = 2 D ; P 2 = 3 D d = 3 1 m We know that F 1 = f 1 1 + f 2 1 − f 1 ⋅ f 2 d Equivatent power, ∴ P = P 1 + P 2 − d P 1 ⋅ P 2 = 2 + 3 − 3 1 × 2 × 3 = 3 D

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PhysicsGeometrical OpticsRefraction+ReflectionMedium2 minPYQ_2015

PhysicsMediumsingle choice

Two convex lenses of power $2 D$ and $5 D$ are separated by a distance $\frac{1}{3} m$ . The power of optical system formed is

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Given: $P_{1} = 2 D; P_{2} = 3 D d = \frac{1}{3} m$ We know that $\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}} - \frac{d}{f _{1} \cdot f _{2}}$ Equivatent power, $∴ P = P_{1} + P_{2} - d P_{1} \cdot P_{2} = 2 + 3 - \frac{1}{3} \times 2 \times 3 = 3 D$

Stream:VITEEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Refraction+Reflection

⏱ 2mℹ️ Source: PYQ_2015

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