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An object in placed 30cm in front of a c...

An object in placed 30cm in front of a concave lens that is made of a glass of refractive index 1.5 and has equal radii of curvature of its two surfaces, each 30 cm. The surface of the lens farther away from the object is silvered. Find the nature and position of the final image.

Text Solution

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Focal length of the concave lens, using the lensmaker's formula, is
`(1)/(f_(L))=(1.5-1)(-(1)/(30)-(1)/(30)) [R_(1)=-30cm,R_(2)=30cm]`
`f_(L)=-30cm=-0.3m`
In a silvered lens, light as it enters the lens suffers refraction, then it gets reflected at the silvered surface, and again undergoes refraction as it comes out in air.
In such a situation, power of the silvered lens will be
`P=P_(L)+P_(M)=2P_(L)+P_(M)`
`P_(L)=(1)/(f_(L))=-(1)/(0.3)D`
`P_(M)=-(1)/(f_(M))=-(1)/(0.15)D`
`[` Silvered surface behaves as a spherical convex mirror of radius of curvature 30 cm so that focal length will be `15 cm =0.15m]`
`P=2P_(L)+P_(M)=-(2)/(0.3)-(1)/(1.5)=-13.33D`
And focal length of the equivalent mirror,
`F=-(1)/(P)-(1)/(-13.33)=0.075m=7.5mm`.
i.e., the silvered lens behaves as a convex mirror of focal length 7.5cm.
For an object placed at a distance of 30 cm from the silvered lens, which behaves as a convex mirror of focal length 7.5cm,
`u=-30 cm` and `f=7.5 cm`
`(1)/(v)+(1)/(u)=(1)/(f)rArr(1)/(v)-(1)/(30)=(1)/(7.5)`
`(1)/(v)-(1)/(7.5)+(1)/(30)`
`rArr v=6cm`
i.e., the image will be 6cm to the right of the silvered lens and it will be a virtual image [ silvered lens behaves as a convex mirror ] .
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