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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsRefraction at curved surface/LensMedium2 minPYQ_2011

PhysicsMediumsingle choice

A conversing beam of rays is incident on a diverging lens. Having passed though the lens the rays intersect at a point $15 cm$ from the lens on the opposite side. If the lens is removed the point where the rays meets will move $5 cm$ closer to the lens. The focal length of the lens is

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Solution:

Given $u = 10 cm, y = 15 cm$ $\frac{1}{f} = \frac{1}{y} - \frac{1}{u} \frac{1}{f} = \frac{1}{15} - \frac{1}{10} \frac{1}{f} = \frac{10 - 15}{150} \frac{1}{f} = \frac{- 5}{150} f = - 30 cm$

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

⏱ 2mℹ️ Source: PYQ_2011

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