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ICSE Class 9 Physics Question 15 of 19

Reflection of Light — Question 11

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Question

Question 11

A concave mirror forms a virtual image of size twice that of the object placed at a distance 5 cm from it. Find (a) the focal length of the mirror (b) position of the image.

Answer

(a) Given,

Distance of the object (u) = 5 cm (negative)

Magnification (m) = 2 (positive for the virtual image)

Magnification (m)=Length of image (I)Length of object (O)=Distance of image (v)Distance of object (u)2=−v−5⇒v=2×5⇒v=10 cm\text{Magnification (m)} = \dfrac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[0.5em] = \dfrac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] 2 = - \dfrac{\text v}{- 5} \\[0.5em] \Rightarrow \text v = 2 \times 5 \\[0.5em] \Rightarrow \text v = 10 \text{ cm} \\[0.5em]

Hence, the image is formed 10 cm behind the mirror.

Mirror formula:

1u+1v=1f\dfrac{1}{\text u} + \dfrac{1}{\text v} = \dfrac{1}{\text f}

Substituting the values in the mirror formula, we get,

−15+110=−1f−1f=−2+110−1f=−110⇒f=10 cm- \dfrac{1}{5} + \dfrac{1}{10} = - \dfrac{1}{\text f} \\[0.5em] - \dfrac{1}{\text f} = \dfrac{ - 2 + 1 }{10} \\[0.5em] - \dfrac{1}{\text f} = -\dfrac{1}{10} \\[0.5em] \Rightarrow \text f = 10 \text{ cm} \\[0.5em]

Hence, focal length of the mirror = 10 cm

(b) The position of the image is 10 cm behind the mirror