A convex mirror has a radius of 20 cm. An object is placed 50 cm from the mirror. Find the location of the image, the magnification, and the type of image formed. 2. For the previous mirror system, draw the three principle rays and compare your results to that obtained for the previous problem. 3. A convex mirror has a radius of 60 cm. An object is placed 200 cm from the mirror. Find the location of the image, the magnification, and the type of image formed. For the previous mirror system, draw the three principle rays and compare your results to that obtained for the previous problem.

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{-10}-\frac{1}{50}\\\Rightarrow \frac{1}{v}=\frac{-3}{25} \\\Rightarrow v=\frac{-25}{3}=-8.33\ cm[/tex]

Image is 8.33 cm behind the mirror

As, the sign is negative the image is behind the mirror and is virtual in nature

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8.33}{50}\\\Rightarrow m=0.166[/tex]

Magnification is 0.166

Positive magnification indicates the image is upright.

R = Radius of curvature = 60 cm

Focal length = f = R/2 = 60/2 = 30 cm

f = -30 cm (convex mirror)

u = Object distance = 200 cm

Lens Equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{-30}-\frac{1}{200}\\\Rightarrow \frac{1}{v}=\frac{-23}{600} \\\Rightarrow v=\frac{-600}{23}=-26.08\ cm[/tex]

Image is 26.08 cm behind the mirror

As, the sign is negative the image is behind the mirror and is virtual in nature

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-26.08}{200}\\\Rightarrow m=0.13[/tex]

Magnification is 0.13

Positive magnification indicates the image is upright.