Question

A 4.5cm needle is placed 12cm away from a convex mirror of focal length 15cm. Give the location of the image and the magnification.Describe what happens as the needle is moved farther from the mirror.

Answer 1

Height of the needle,h1=4.5cm
Object distance,u==-12cm
The focal length of the convex mirror,f=15cm
Image distance=v
The value of v can be obtained using the mirror formula:\(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)
\(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\)=\(\frac{1}{15}+\frac{1}{12}\)=\(4+\frac{5}{60}\)=\(\frac{9}{60}\)
∴ v=\(\frac{60}{9}\)=6.7cm
Hence, the image of the needle is 6.7cm away from the mirror. Also, it is on the other side of the mirror. The image is given by the magnification formula:m=\(\frac{h_2}{h_1}\)=\(-\frac{v}{u}\)
∴\(h_2=-\frac{v}{u}\times h_1\)=\(\frac{-6.7}{-12}\times4.5\)=+2.5cm
Hence, magnification of the image,m=\(\frac{h_2}{h_1}=\frac{2.5}{4.5}\)=0.56
The height of the image is 2.5cm. The positive sign indicates that the image is erect, virtual, and diminished. If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.