an erect image three times magnified the size of the object is obtained with concave mirror of radius of curvature 36 cm. what is the position of the object?
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ans. 12cm
Explanation:
according to the question
magnification (m)=3 i.e -v/u =3 => v=–3u
by mirror formula : 1/v +1/u =1/f
ATQ– f=18cm
applying formula
–1/3u + 1/u = –1/18 ( focal length of a concave lens is -ve)
=> u= –12cm( – for sign convention)
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