An object 5 cm is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position, nature, and size of the image.

Asked by Vishal kumar | 1 year ago |  369

##### Solution :-

Object distance (u) = - 20 cm

Object height (h) = 5 cm

Radius of curvature (R) = 30 cm

Radius of curvature = 2 × Focal length

R = 2f

f = 15 cm

According to the mirror formula,

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$

$$\frac{1}{u}=\frac{1}{f}+\frac{1}{u}$$

$$=\frac{1}{15}+\frac{1}{20}=\frac{4+3}{60}=\frac{7}{60}$$

v = 8.57 cm

The positive value of v indicates that the image is formed behind the mirror

$$Magnification,m=-\frac{Image \,Distance}{Object \,Distance}=\frac{-8.57}{-20}=0.428$$

The positive value of magnification indicates that the image formed is virtual

$$Magnification,m=\frac{Height \,of\,the\,image}{Height\,of\,the\,object}=\frac{h^1}{h}$$

$$h^1=m\times h=0.428\times5=2.14\,cm$$

The positive value of image height indicates that the image formed is erect. Hence, the image formed is erect, virtual, and smaller in size.

Answered by Shivani Kumari | 1 year ago

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