An object is placed on the principal axis of a concave mirror at a distance of 1.5 f (f is the focal length). The image will be at,

Question

An object is placed on the principal axis of a concave mirror at a distance of 1.5 f (f is the focal length). The image will be at, (A) -3 f (B) 1.5 f (C) -1.5 f (D) 3 f

Tardigrade Admin · Accepted Answer

u = -1.5 f (1/u)+(1/v)=(1/f) =(1/-1.5 f)+(1/v)=(1/-f) ⇒ (1/v)=-(1/f)+(1/1.5 f) (1/v)=(-1.5+1/1.5 f) =(-0.5/1.5 f) v=-3f

Q. An object is placed on the principal axis of a concave mirror at a distance of $1.5 f$ ( $f$ is the focal length). The image will be at,

$- 3 f$

$1.5 f$

$- 1.5 f$

$3 f$

$u = - 1.5 f$
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$= \frac{1}{- 1.5 f} + \frac{1}{v} = \frac{1}{- f}$
$\Rightarrow \frac{1}{v} = - \frac{1}{f} + \frac{1}{1.5 f}$
$\frac{1}{v} = \frac{- 1.5 + 1}{1.5 f}$
$= \frac{- 0.5}{1.5 f}$
$v = - 3 f$

Q. An object is placed on the principal axis of a concave mirror at a distance of 1.5f (f is the focal length). The image will be at,

−3f

1.5f

−1.5f

3f