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ICSE Class 9 Physics Question 10 of 19

Reflection of Light — Question 6

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Question 6

An object 5 cm high is placed at a distance 60 cm in front of a concave mirror of focal length 10 cm. Find (i) the position and (ii) size, of the image.

Answer

Given,

Object height (O) = 5 cm

focal length (f) = 10 cm (negative)

Object distance (u) = 60 cm (negative)

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,

−160+1v=−1101v=−110+1601v=−6+1601v=−5601v=−112v=−12 cm- \dfrac{1}{60} + \dfrac{1}{\text v} = - \dfrac{1}{10} \\[0.5em] \dfrac{1}{\text v} = - \dfrac{1}{10} + \dfrac{1}{60} \\[0.5em] \dfrac{1}{\text v} = \dfrac{ - 6 + 1}{60} \\[0.5em] \dfrac{1}{\text v} = - \dfrac{ 5}{60} \\[0.5em] \dfrac{1}{\text v} = - \dfrac{1}{12} \\[0.5em] \text v = - 12 \text{ cm} \\[0.5em]

The image distance (v) = 12 cm infront of the mirror.

Magnification (m)=Length of image (I)Length of object (O)=Distance of image (v)Distance of object (u)I5=−1260I=−12×560⇒I=−1 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] \dfrac{\text I}{5} = - \dfrac{12}{60} \\[0.5em] \text I = -\dfrac{12 \times 5}{60} \\[0.5em] \Rightarrow \text I = - 1 \text { cm}\\[0.5em]

Hence, length of image = 1 cm

Negative sign shows that the image will be inverted.