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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, 01:00: PM
answered-by-expert Expert Answer

Lens Equation is

img

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

img

img

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, 02:48: PM
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