ICSE Class 9 Physics Question 12 of 19

Reflection of Light — Question 8

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Question 8

When an object of height 1 cm is kept at a distance 4 cm from a concave mirror, it's erect image of height 1.5 cm is formed at a distance 6 cm behind the mirror. Find the focal length of the mirror.

Answer

Given,

u = 4 cm (negative)

v = 6 cm (positive)

Mirror formula:

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

Substituting the values in the formula above, we get,

14+16=1f1f=3+2121f=112f=12 cm- \dfrac{1}{4} + \dfrac{1}{6} = \dfrac{1}{\text f} \\[0.5em] \dfrac{1}{\text f} = \dfrac{-3+2}{12} \\[0.5em] \dfrac{1}{\text f} = - \dfrac{1}{12} \\[0.5em] \Rightarrow \text f = - 12 \text{ cm} \\[0.5em]

Hence, the focal length of concave mirror = 12 cm