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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsRefraction at curved surface/LensMedium2 minQB

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A thin lens of index of refraction 1.5 has one convex side with a radius of magnitude 20 cm. When an object 1 cm in height is placed 50 cm from this lens, an upright image 2 cm in height is formed. Which of the following figures correctly represent the lens?

Options:

Answer:

Solution:

$m = \frac{h _{I}}{h _{O}} = + 2 = \frac{v}{u} \Rightarrow v = 2 u$ (upright image). With u = -50 cm, v = -100 cm: $\frac{1}{- 100} + \frac{1}{50} = \frac{1}{f} \Rightarrow f = 100 cm$ . Then $\frac{1}{100} = (1.5 - 1) (\frac{1}{20} - \frac{1}{R _{2}}) \Rightarrow \frac{1}{20} - \frac{1}{R _{2}} = \frac{1}{50} \Rightarrow R_{2} = \frac{100}{3} cm$ . Both surfaces are convex to the object side with the second surface concave as seen from the other side: a converging meniscus. Hence (B).

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Refraction at curved surface/Lens

⏱ 2mℹ️ Source: QB

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