Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsRefraction+ReflectionEasy2 minPYQ_2020

PhysicsEasysingle choice

A thin convex lens $L$ (refractive index $= 1.5$ ) is placed on a plane mirror $M$ . When a pin is placed at $A$ , such that $O A = 18 c m,$ its real inverted image is formed at $A$ itself, as shown in figure. When liquid of refractive index $μ_{l}$ is put between the lens and the mirror, the pin has to be moved to $A^{'},$ such that $O A^{'} = 27 c m,$ to get its inverted real image at $A'$ itself. The value of $μ_{l}$ will be

Options:

Answer:

Solution:

$\frac{1}{f_{1}} = \frac{1}{2} \times \frac{2}{18}$
$\frac{1}{f_{2}} = \frac{(μ_{1} - 1)}{- 18}$
When $μ_{1}$ is filled between lens and mirror,
$P = \frac{2}{18} - \frac{2}{18} (μ_{1} - 1) = \frac{2 - 2 μ_{1} + 2}{18}$
$\Rightarrow F_{m} = - (\frac{18}{2 - μ_{1}})$
$2 = 6 - 3 μ_{1}$
$3 μ_{1} = 4$
$μ_{1} = \frac{4}{3}$

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Refraction+Reflection

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...