A convex mirror has a focal length f. A real object is placed at a distance f in front of it, from the pole. It produces an image at

Question

A convex mirror has a focal length f. A real object is placed at a distance f in front of it, from the pole. It produces an image at (A) Infinity (B) f (C) f / 2 (D) 2 f

Tardigrade Admin · Accepted Answer

Mirror formula: (1/v)+(1/u)=(1/f) Here object is real so u is negative (1/v)-(1/u)=(1/f) Also (u)=f (1/v)-(1/f)=(1/f) ∴ v=(f/2)

Q. A convex mirror has a focal length $f$ . A real object is placed at a distance $f$ in front of it, from the pole. It produces an image at

Infinity

f

f / 2

2 f

Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
Here object is real so $u$ is negative
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Also $(u) = f$
$\frac{1}{v} - \frac{1}{f} = \frac{1}{f}$
$∴ v = \frac{f}{2}$

Q. A convex mirror has a focal length f. A real object is placed at a distance f in front of it, from the pole. It produces an image at

Infinity

f

f / 2

2 f

Mirror formula: v1​+u1​=f1​ Here object is real so u is negative v1​−u1​=f1​ Also (u)=f v1​−f1​=f1​ ∴v=2f​

Q. A convex mirror has a focal length $f$ . A real object is placed at a distance $f$ in front of it, from the pole. It produces an image at

Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
Here object is real so $u$ is negative
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Also $(u) = f$
$\frac{1}{v} - \frac{1}{f} = \frac{1}{f}$
$∴ v = \frac{f}{2}$