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ICSE Class 10 Physics Question 2 of 27

Solved Sample Paper 3 — Question 12

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12
Question

Question 1(xii)

What is the magnification of the image when a lens of focal length 10 cm and object distance from the lens is 15 cm?

  1. 2
  2. 1
  3. 12\dfrac{1}{2}
  4. 4
Answer

2

Reason — As we know, the lens formula is —

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

Given,

f = 10 cm

u = -15 cm

v = ?

Substituting the values in the formula, we get,

110=1v–1−15110=1v+1151v=110−1151v=3−2301v=130⇒v=30 cm\dfrac{1}{10} = \dfrac{1}{v} – \dfrac{1}{-15} \\[1em] \dfrac{1}{10} = \dfrac{1}{v} + \dfrac{1}{15} \\[1em] \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15} \\[1em] \dfrac{1}{v} = \dfrac{3-2}{30} \\[1em] \dfrac{1}{v} = \dfrac{1}{30} \\[1em] \Rightarrow v = 30 \text{ cm} \\[1em]

∴ Image distance is 30 cm.

As we know, the formula for magnification of a lens is —

m=vum = \dfrac{v}{u}

Given,

v = 30 cm

u = -15 cm

Substituting the values in the formula, we get,

m=30−15m=−2m = \dfrac{30}{- 15} \\[1em] m = -2

∴ The magnification is 2, negative sign implies image is inverted.