Answer:
1.Theimage will be located at -0.13m or -13 cm
2.The height of the image will be 0.052m or 5.2cm
Explanation:
Given that;
Height of object, h=20 cm = 0.2m
Object distance in front of convex mirror, o,= 50 cm =0.5m
Radius of curvature, r, =34 cm =0.34m
Let;
Image distance, i,=?
Image height, h'=?
You know that focal length,f, is half the radius of curvature,hence
f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)
f= -0.17m
Apply the relationship that involves the focal length;
![=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7Bo%7D%20%2B%5Cfrac%7B1%7D%7Bi%7D%20%3D%5Cfrac%7B1%7D%7Bf%7D)
![=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B0.5%7D%20%2B%5Cfrac%7B1%7D%7Bi%7D%20%3D-%5Cfrac%7B1%7D%7B0.17%7D)
Re-arrange to get i
![\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bi%7D%20%3D-2-5.88%5C%5C%5C%5C%5C%5C%5Cfrac%7B1%7D%7Bi%7D%20%3D-7.88%5C%5C%5C%5Ci%3D-0.13m)
This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror
Apply the magnification formula
magnification, m=height of image÷height of object
![m=\frac{h'}{h} =-\frac{i}{o}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7Bh%27%7D%7Bh%7D%20%3D-%5Cfrac%7Bi%7D%7Bo%7D)
substitute the values to get the height of image h'
![\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm](https://tex.z-dn.net/?f=%5Cfrac%7Bh%27%7D%7B0.20%7D%20%3D-%5Cfrac%7B-0.13%7D%7B0.5%7D%20%5C%5C%5C%5C%5C%5Ch%27%3D%5Cfrac%7B0.13%2A0.20%7D%7B0.5%7D%20%5C%5C%5C%5C%5C%5Ch%27%3D%5Cfrac%7B0.025%7D%7B0.5%7D%20%3D0.052m%5C%5C%5C%5C%5C%5Ch%27%3D5.2cm)