An object is placed at 10 cm in front of a concave mirror of radius of curvature 15 cm. The magnification and nature of the image is

Question

An object is placed at 10 cm in front of a concave mirror of radius of curvature 15 cm. The magnification and nature of the image is (A) 2 times, real, inverted (B) 2,times, virtual, erect (C) 3 times, real, inverted (D) 3 times, virtual, erect

Tardigrade Admin · Accepted Answer

f=(R/2)=(-15/2)=-7.5 cm and u=-10 cm , ∴ (1/v)+(1/(-10))=(1/-7.5) ⇒ v=-30 cm The -ve value of v, means image is real. m=-(v/u)=(-(30)/-10)=-3 The image is inverted and magnified 3 times.

Q. An object is placed at 10 cm in front of a concave mirror of radius of curvature 15 cm. The magnification and nature of the image is

2 times, real, inverted

2,times, virtual, erect

3 times, real, inverted

3 times, virtual, erect

$f = \frac{R}{2} = \frac{- 15}{2} = - 7.5 c m$
and $u = - 10 c m, ∴ \frac{1}{v} + \frac{1}{( - 10 )} = \frac{1}{- 7.5}$
$\Rightarrow v = - 30 c m$
The -ve value of $v,$ means image is real.
$m = - \frac{v}{u} = \frac{- ( 30 )}{- 10} = - 3$
The image is inverted and magnified 3 times.

Q. An object is placed at 10 cm in front of a concave mirror of radius of curvature 15 cm. The magnification and nature of the image is

2 times, real, inverted

2,times, virtual, erect

3 times, real, inverted

3 times, virtual, erect

f=2R​=2−15​=−7.5cm and u=−10cm,∴v1​+(−10)1​=−7.51​ ⇒v=−30cm The -ve value of v, means image is real. m=−uv​=−10−(30)​=−3 The image is inverted and magnified 3 times.

$f = \frac{R}{2} = \frac{- 15}{2} = - 7.5 c m$
and $u = - 10 c m, ∴ \frac{1}{v} + \frac{1}{( - 10 )} = \frac{1}{- 7.5}$
$\Rightarrow v = - 30 c m$
The -ve value of $v,$ means image is real.
$m = - \frac{v}{u} = \frac{- ( 30 )}{- 10} = - 3$
The image is inverted and magnified 3 times.