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,$m=0.25$

Vedclass · Answer

(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 $(m) = -\frac{v}{u} = -\frac{0.375}{-1.5} = \frac{0.375}{1.5} = 0.25$.

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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