Consider a light source placed at a distance | vedclass

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(A) Given: Distance of light source $(u) = -1.5 \ m = -\frac{3}{2} \ m$. Radius of curvature of the convex mirror $(R) = +1 \ m$. Focal length $(f) =

Vedclass · Accepted Answer

$s^{\prime}=0.375 \ m$, virtual, upright, $m=0.25$

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(A) Given: Distance of light source $(u) = -1.5 \ m = -\frac{3}{2} \ m$. Radius of curvature of the convex mirror $(R) = +1 \ m$. Focal length $(f) = \frac{R}{2} = +0.5 \ m$.Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Substituting the values: $\frac{1}{0.5} = \frac{1}{v} + \frac{1}{-1.5} \Rightarrow 2 = \frac{1}{v} - \frac{2}{3}$.Solving for $v$: $\frac{1}{v} = 2 + \frac{2}{3} = \frac{6+2}{3} = \frac{8}{3}$.Therefore, $v = \frac{3}{8} = 0.375 \ m$.Since $v$ is positive, the image is formed behind the mirror, making it virtual and upright.Magnification $(m) = -\frac{v}{u} = -\frac{0.375}{-1.5} = \frac{0.375}{1.5} = 0.25$.

Column-$I$	Column-$II$
$1$. $m = -2$	a. Convex mirror
$2$. $m = -1/2$	b. Concave mirror
$3$. $m = +2$	c. Real image
$4$. $m = +1/2$	d. Virtual image

Consider a light source placed at a distance of $1.5 \ m$ along the axis facing the convex side of a spherical mirror of radius of curvature $1 \ m$. The position $(s^{\prime})$, nature and magnification $(m)$ of the image are

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Match the corresponding entries of Column-$I$ with Column-$II$. [Where $m$ is the magnification produced by the mirror]
Column-$I$ Column-$II$
$1$. $m = -2$ a. Convex mirror
$2$. $m = -1/2$ b. Concave mirror
$3$. $m = +2$ c. Real image
$4$. $m = +1/2$ d. Virtual image

$A$ straight line joining the object point and image point is always perpendicular to the mirror:

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