a numerical :lenses
Asked by | 29th Jan, 2009, 10:33: AM
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 | 30th Jan, 2009, 10:50: AM
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