Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsRefraction at curved surface/LensEasy2 minPYQ_2020

PhysicsEasynumerical

Diameter or aperture of a plano-convex lens is $6 c m$ and its thickness at the centre is $3 mm.$ The image of an object formed is real and twice the size of the object. If the speed of light in the material of the lens is $2 \times 10^{8} {m s}^{- 1}$ . The distance where the object is placed from the plano-convex lens is $\dots \dots \dots \times 15 c m$ .

Answer:

3.00

Solution:

${(R - y)}^{2} + r^{2} = R^{2}$

$R = \frac{r^{2}}{2 y} = \frac{{(\frac{6}{2})}^{2}}{2 \times 0 .3} = 15 cm$

$\frac{1}{f} = \frac{μ - 1}{R}$

$μ = \frac{C}{V} = \frac{3 .0 \times 10^{8}}{2 .0 \times 10^{8}} = 1 .5$

$∴ f = 30 cm$ converging

Now

$f = 30 cm$

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

$\frac{1}{30} = \frac{1}{v} - \frac{1}{u}$

$m = \frac{v}{u} = - 2$

$v = - 2 u$

$\frac{1}{30} = \frac{- 1}{2 u} - \frac{1}{u}$

$\frac{1}{30} = - \frac{3}{2 u}$

$u = - 45 cm$

Subject:PhysicsTopic:Geometrical OpticsSubtopic:Refraction at curved surface/Lens

⏱ 2mℹ️ Source: PYQ_2020

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