A biconvex lens of focal length 40 cm is placed in front of an object at a distance of 20 cm . Now a slab of refractive index 4 3 is placed somewhere in between the lens and the object. The shift in the image formed after the introduction of slab equals (thickness of the slab is 2 mm )

Before the slab is introduced, f = 40 cm , u = - 10 cm 1 v - 1 u = 1 f ⇒ 1 v + 1 20 = 1 40 ⇒ v = - 40 c m Shift in image of object due to slab = t 1 - 1 μ S = 2 1 - 3 4 = 1 2 m m = 0.5 m m = d u Also, d v d u = v 2 u 2 d v = v 2 u 2 d u 40 20 2 d u = 4 × 0.5 m m = 2 m m

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PhysicsGeometrical OpticsRefraction at curved surface/LensEasy2 minPYQ_2020

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A biconvex lens of focal length $40 cm$ is placed in front of an object at a distance of $20 cm$ . Now a slab of refractive index $\frac{4}{3}$ is placed somewhere in between the lens and the object. The shift in the image formed after the introduction of slab equals (thickness of the slab is $2 mm$ )

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Answer:

Solution:

Before the slab is introduced,

$f = 40 cm, u = - 10 cm$

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{1}{v} + \frac{1}{20} = \frac{1}{40}$

$\Rightarrow v = - 40 c m$

Shift in image of object due to slab $= t (1 - \frac{1}{μ_{S}})$

$= 2 (1 - \frac{3}{4})$

$= \frac{1}{2} m m = 0.5 m m = d u$

Also, $\frac{d v}{d u} = \frac{v^{2}}{u^{2}}$

$d v = \frac{v^{2}}{u^{2}} d u$

${(\frac{40}{20})}^{2} d u = 4 \times 0.5 m m$

$= 2 m m$

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

⏱ 2mℹ️ Source: PYQ_2020

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