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Physics

An object is placed at a distance 24 cm in front of a convex lens of focal length 8 cm.

(a) What is the nature of the image so formed?

(b) Calculate the distance of the image from the lens.

Refraction Lens

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Answer

(a) As, the object distance (= 24 cm) is greater than the focal length (= 8 cm) of the convex lens which means the object placed beyond it's focus so the lens will form a real and inverted image.

(b)

Given,

Object distance (u) = - 24 cm

Focal length (f) = +8 cm

From lens formula,

1f=1v−1u1v=1f+1u1v=18+1(−24)[1em]1v=18−1241v=3−1241v=224v=242v=12 cm\dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{\text u} \\[1 em] \dfrac{1}{\text v} = \dfrac{1}{\text f} + \dfrac{1}{\text u} \\[1 em] \dfrac{1}{\text v} = \dfrac{1}{8}+\dfrac{1}{(-24)} \\ [1 em] \dfrac{1}{\text v} =\dfrac{1}{8}-\dfrac{1}{24} \\[1 em] \dfrac{1}{\text v} = \dfrac{3-1}{24}\\[1 em] \dfrac{1}{\text v} = \dfrac{2}{24} \\[1 em] \text v = \dfrac{24}{2} \\[1 em] \text v =12 \text { cm}

∴ The image is formed at a distance of 12 cm behind the lens.

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