Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsReflection at curved surfaceEasy2 minPYQ_2020

PhysicsEasysingle choice

When an object is placed at a distance of $25 cm$ from a mirror, the magnification is $m_{1}$ . But the magnification becomes $m_{2}$ , when the object is moved $15 cm$ farther away with respect to the earlier position. $\frac{m_{1}}{m_{2}} = 4,$ then find the focal length of the mirror and what type of mirror it is?

Options:

Answer:

Solution:

$m = - \frac{v}{u} = \frac{f}{f - u}$

$so, m_{1} = \frac{f}{f - (- 25)} = \frac{f}{f + 25}$

$Similarly, m_{2} = \frac{f}{f + 40}$

$and \frac{m_{1}}{m_{2}} = 4$

$∴ \frac{f + 40}{f + 25} = 4$

$f = - 20 cm$

negative sign shows that the mirror is concave.

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Geometrical OpticsSubtopic:Reflection at curved surface

⏱ 2mℹ️ Source: PYQ_2020

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