Question

A convex lens forms an image 16.0 cm long of an object 4.0 cm long kept at a distance of 6 cm from the lens. The object and the image are on the same side of the lens. (a) What is nature of the image? (b) Find (i) the position of the image, and (ii) the focal length of the lens. [Need quality answers from best users of Brainly]​

Answer 1

Answer:

(a). real image

(b) beyond the centre of curvature

Answer 2

Answer:

hey here is your solution

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Explanation:

$To \: find = \\ 1.nature \: of \: image \\ 2.position \: of \: image \\ 3.focal \: length \: (f) \: of \:the \: lens \\ \\ Given = \\ for \: a \: convex \: lens\\ height \: of \: image \: (h2) = 16.cm \\ height \: of \: object \: (h1) = 4.cm \\ object \: distnce \: (u) = - 6.cm \\ \\ so \: we \: know \: that \\ \frac{h2}{h1} = \frac{v}{u} \\ \\ \frac{16}{4} = \frac{v}{ - 6} \\ \\ 4 = \frac{v}{ - 6} \\ \\ = 4 \times ( - 6) \\ hence \: v = - 24.cm \\ \\ so \: as \: image \: distance \: is \: negative \\ the \: image \: formed \: is \: virtual \: and \: erect \\ \\ and \: we \: know \: that \\ convex \: lens \: only \: forms \: virtual \: and \: erect \: image \: \\ when \: it \: is \: placed \: between \: optical \: centre \: O \: and \: focus \: F1$

$now \: applying \: lens \: formula \\ we \: get \\ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \\ = \frac{1}{( - 24)} \: - \: \frac{1}{( - 6)} \\ \\ = \frac{1}{( - 24)} + \frac{1}{6} \\ \\ = \frac{6 - 24}{6 \times ( - 24)} \\ \\ = \frac{ - 18}{ - 144} \\ \\ = \frac{18}{144} \\ \\ = \frac{1}{8} \\ \\ ie \: \: \frac{1}{f} = \frac{1}{8} \\ \\ so \: f = 8.cm \\ \\ hence \: focal \: length \: f \: is \: 8.cm$