An object of size 3.0" " cm is placed in front of a concave mirror of focal length 18" " cm, at a distance of 54" " cm. The image formed is and its height is

Correct option is A

Given:

Focal length $$f$$ = -18 cm (Since it is concave mirror, the focal length is negative)

Object distance $$u$$ = -54cm (Since the object is placed in front of the mirror, the object distnace is negative)

Object height $$h_o = 3.0 \, \text{cm}$$​​

Formula used:

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

Solution:

​$$f = -18 \, \text{cm}, \quad u = -54 \, \text{cm}$$ 

$$\frac{1}{-18} = \frac{1}{v} + \frac{1}{-54}$$ 

​$$v = -27 \, \text{cm}$$ 

$$m = \frac{v}{u} = \frac{-27}{-54} = \frac{1}{2}$$​​

​$$h_i = m \times h_o = \frac{1}{2} \times 3.0 = 1.5 \, \text{cm}$$ 

Thus the correct answer is (A)

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