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CBSE Class 10 Answered

a numerical :lenses
Asked by | 29 Jan, 2009, 10:33: AM
answered-by-expert Expert Answer

Sorry, for the answer to this question last time. I had mentioned that the image will be virtual, but since the silvered side is concave a real image can be formed.

In this case refraction takes place when light enters the lens, reflection from the curved surface and then refraction when light emerges from the lens.

1/F = 1/f + 1/fm + 1/f = 2/f +1/fm

fm is the focal length of the mirror = R/2.----(1)

f is the focal length of the curved surface which is found using 1/f = (μ-1)(1/R) so f = R/(μ-1)----(2)

Hence effective focal length using 1 and 2 is 1/F = 2/f + 1/fm 

F = R/2μ = 30/2(1.5) = 10cm.

So if the object is at 20 cm a real image of the same size as the object is formed.

Answered by | 30 Jan, 2009, 10:50: AM
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