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ICSE Class 9 Physics Question 8 of 19

Reflection of Light — Question 3

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Question 3

An object of height 2 cm is placed at a distance 20 cm in front of a concave mirror of focal length 12 cm. Find the position, size and nature of the image.

Answer

Given,

Object height (O) = 2 cm

Focal length (f) = 12 cm (negative)

Object distance (u) = 20 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,

−120+1v=−1121v=−112+1201v=−5+3601v=−2601v=−130v=−30 cm- \dfrac{1}{20} + \dfrac{1}{\text v} = - \dfrac{1}{12} \\[0.5em] \dfrac{1}{\text v} = - \dfrac{1}{12} + \dfrac{1}{20} \\[0.5em] \dfrac{1}{\text v} = \dfrac{ - 5 + 3 }{60} \\[0.5em] \dfrac{1}{\text v} = - \dfrac{ 2}{60} \\[0.5em] \dfrac{1}{\text v} = - \dfrac{1}{30} \\[0.5em] \text v = - 30 \text{ cm} \\[0.5em]

Hence, the image is formed at a distance of 30 cm in front of the mirror.

Magnification (m)=Length of image (I)Length of object (O)=Distance of image (v)Distance of object (u)⇒I2=−3020⇒I=−30×220⇒I=−3 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] \Rightarrow \dfrac{\text I}{2} = - \dfrac{30}{20} \\[0.5em] \Rightarrow \text I = -\dfrac{30 \times 2}{20} \\[0.5em] \Rightarrow \text I = - 3 \text { cm}\\[0.5em]

Hence, length of image = 3 cm

Image will be real, inverted and magnified.