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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsRefraction at curved surface/LensMedium3 minQB

PhysicsMediuminteger

The focal length of a thin biconvex lens is 20cm. When an object is moved from a distance of 25cm in front of it to 50cm, the magnification of its image changes from $m_{25}$ to $m_{50}$ . The ratio $\frac{m _{25}}{m _{50}}$ is

Answer:

Solution:

$u = 25 : \frac{1}{v} = \frac{1}{20} - \frac{1}{25} \Rightarrow v = 100 cm$ , $m_{25} = \frac{100}{25} = 4$

$u = 50 : \frac{1}{v} = \frac{1}{20} - \frac{1}{50} \Rightarrow v = \frac{100}{3} cm$ , $m_{50} = \frac{100/3}{50} = \frac{2}{3}$

$\frac{m _{25}}{m _{50}} = \frac{4}{2/3} = 6$

Answer: 6

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

⏱ 3mℹ️ Source: QB

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