A coin of diameter 10 cm is placed at a distance 3.5 f from the pole of a concave mirror of focal length f. The linear magnification and the diameter of the image are:

Correct option is A

The correct answer is (A)  $$\frac{2}{5}$$ ​and 4 cm, respectively

Given:

• Diameter of coin($$h_o$$​) = 10 cm
  
• Object distance(u) = - 3.5 f
• Focal length (f) = f

Formula used:   Mirror formula:

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

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{3.5f}$$ 

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{3.5f} = \frac{3.5 - 1}{3.5f} = \frac{2.5}{3.5f}$$

v = $$\frac{3.5f}{2.5} = 1.4f$$

Linear magnification (m): 

m = $$\frac{-v}{u} = -\frac{1.4f}{-3.5f} = \frac{1.4}{3.5} = 0.4 = \frac{2}{5}$$

Diameter of image($$h_i$$​): 

$$h_i$$ = m.$$h_o$$ = 0.4 . 10cm = 4cm

Linear magnification (m):$$\frac{2}{5}$$​

Diameter of image: 4cm