Question 9An object of length 4 cm is placed in front of a concave mirror at a distance 30 cm. The focal length of mirror is 15 cm. (a) Where will the image form? (b) What will be the length of image?

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Accepted Answer

(a) Given,u = 30 cm (negative)f = 15 cm (negative)Mirror formula:1u+1v=1f\dfrac{1}{\text u} + \dfrac{1}{\text v} = \dfrac{1}{\text f}u1​+v1​=f1​Substituting the values in the formula above, we get,−130+1v=−1151v=−115+1301v=−2+1301v=−130⇒v=30 cm- \dfrac{1}{30} + \dfrac{1}{\text v} = - \dfrac{1}{15} \[0.5em] \dfrac{1}{\text v} = - \dfrac{1}{15} + \dfrac{1}{30} \[0.5em] \dfrac{1}{\text v} = \dfrac{ - 2 + 1 }{30} \[0.5em] \dfrac{1}{\text v} = -\dfrac{1}{30} \[0.5em] \Rightarrow \text v = 30 \text{ cm} \[0.5em]−301​+v1​=−151​v1​=−151​+301​v1​=30−2+1​v1​=−301​⇒v=30 cmHence, the image is formed at 30 cm in front of the mirror(b) MagnificationMagnification (m)=Length of image (I)Length of object (O)=Distance of image (v)Distance of object (u)I4=−3030I4=−1⇒I=−4 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}{4} = - \dfrac{30}{30} \[0.5em] \dfrac{\text I}{4} = - 1 \[0.5em] \Rightarrow \text I = - 4 \text { cm}\[0.5em]Magnification (m)=Length of object (O)Length of image (I)​=Distance of object (u)Distance of image (v)​4I​=−3030​4I​=−1⇒I=−4 cmHence, length of image = 4 cmNegative sign represents that the image is inverted.

Reflection of Light — Question 9

Question 9