CBSE Class 12-science Answered
A tree is 18.0 m away and 2.0 m high from a concave lens. How high is the image formed by the given lens of focal length 6 m ?
Ans is Focal length of the lens is f = -6.0 m,
u = -18 m and h = 2 m
?1v=1f+1u=1(?6.0)?118.0?v=?4.5m
? The image size, h'= h [?vu]=2×[??4.518.0]=0.50m
My doubt is the magnification is v/u for lens right?
Asked by alnaasha2006 | 23 May, 2024, 13:00: PM
Lens Equation is
where v = lens-to-image distance , u = -18 m is lens-to-object distance and f = - 6 m is focal length of caoncave lens.
By cartesian convention lens-to-object distance is negative and focal length is also negative .
Hence we calculated lens-to-image distance v using lens equation as
Hence v = -4.5 m , i.e. image is formed at same side of object at a distance 4.5 m from lens.
Magnification m = v/u = (-4.5)/(-18) = (1/4)
Height of image = (1/4) × 2 m = 0.5 m
Answered by Thiyagarajan K | 23 May, 2024, 14:48: PM
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