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ICSE Class 10 Physics Question 11 of 20

Solved 2019 Question Paper ICSE Class 10 Physics — Question 6

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6
Question

Question 6(c)

An object is placed at a distance 24 cm in front of a convex lens of focal length 8 cm. [4]

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

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

(iii) Calculate the magnification of the image.

Answer

Given u = 24 cm (-ve), f = 8 cm (+ve)

(i) Image formed is real, inverted and diminished.

(ii) Let v be the distance of image from optical centre. From the relation,

1v−1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]

Substituting the values we get,

1v+124=181v=18−1241v=3−1241v=2241v=112⇒v=12cm\dfrac{1}{v} + \dfrac{1}{24} = \dfrac{1}{8} \\[0.5em] \dfrac{1}{v} = \dfrac{1}{8} - \dfrac{1}{24} \\[0.5em] \dfrac{1}{v} = \dfrac{3 - 1}{24} \\[0.5em] \dfrac{1}{v} = \dfrac{2}{24} \\[0.5em] \dfrac{1}{v} = \dfrac{1}{12} \\[0.5em] \Rightarrow v = 12 \text{cm}

Hence, the distance of the image from the lens = 12 cm

(iii) From relation, m = vu\dfrac{v}{u}

Substituting the values we get,

m=12−24⇒m=−0.5m = \dfrac{12}{-24} \\[0.5em] \Rightarrow m = -0.5

-ve sign of magnification tells us that the image is inverted.

Hence, magnification = -0.5