Respuesta :
Answer:
Position = [tex]\frac{60}{7}\ cm[/tex] behind the mirror
Nature = Virtual and Erect
Size = [tex]\frac{15}{7}\ cm[/tex] : Diminished
Explanation:
Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.
Object distance = u = -20 cm
Focal length = f = Radius of curvature/2 = 30/2 = 15 cm
We have to use mirror formula to find image distance.
[tex]\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\\ \frac{1}{-20}+\frac{1}{v}=\frac{1}{15}\\ \frac{1}{v}=\frac{7}{60}\\v=\frac{60}{7}\ cm[/tex]
Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.
Magnification = [tex]=\frac{h_{image}}{h_{object}}=[/tex][tex]-\frac{v}{u}=\frac{60}{7\times20}=\frac{3}{7}[/tex]
Height of the object = 5 cm
Height of the image = [tex]5\times\frac{3}{7}=\frac{15}{7}\ cm[/tex]
Since the height of the image is positive and less than the size of object,it is erect and diminished.
