KnowledgeBoat Logo

Physics

An object is placed at a distance of 12 cm from a convex lens of focal length 8 cm.

Find:

(i) the position of the image

(ii) nature of the image

Refraction Lens

ICSE 2018

12 Likes

Answer

(i) Given,

u = 12 cm (negative),

f = 8 cm (positive)

From lens formula

1v−1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}

Substituting the values we get,

1v−1−12=18⇒1v+112=18⇒1v=18−112⇒1v=3−224⇒1v=124⇒v=24 cm\dfrac{1}{v} - \dfrac{1}{-12} = \dfrac{1}{8} \\[0.5em] \Rightarrow \dfrac{1}{v} + \dfrac{1}{12} = \dfrac{1}{8} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{8} - \dfrac{1}{12} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{3 - 2}{24} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{24} \\[0.5em] \Rightarrow v = 24 \text{ cm}\\[0.5em]

The image will be formed at a distance 24 cm behind the lens.

(iii) The image will be real, inverted and magnified.

Answered By

7 Likes

Related Questions