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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsOptical instrumentsMedium2 minQB

PhysicsMediumsingle choice

The least distance of distinct vision for a farsighted person is 1 m. The optical power of the lens of his specs which effectively reduces his LDDV to 25 cm, is

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Solution:

$- \frac{1}{u} + \frac{1}{v} = \frac{1}{f}$ with $u = - 25 cm, v = - 100 cm$ : $\frac{1}{25} - \frac{1}{100} = \frac{1}{f} \Rightarrow$ power $= 3 D$ .

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Optical instruments

⏱ 2mℹ️ Source: QB

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