# Teacher I was having a doubt regarding this question. So can u please explain me in detailed . As my exam's are going to start so can u explain me in a easy and fast way

The question is of geometric optics and can be solved using the standard mirror formula for concave mirrors, which can be stated as follows:

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

Given,

the object distance, u = -25 cm ( since, the object is on the left side of the mirror)

Focal length of the mirror, f = -15 cm

Thus, its image distance (v) would be:

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\frac{1}{-25}+\frac{1}{v}=\frac{1}{-15}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=\frac{1}{-15}+\frac{1}{25}=\frac{-5+3}{75}=-\frac{2}{75}\phantom{\rule{0ex}{0ex}}Thus,\phantom{\rule{0ex}{0ex}}v=-37.5cm\phantom{\rule{0ex}{0ex}}Thus,theimageisonthesamesideasthatoftheobject.\phantom{\rule{0ex}{0ex}}Itisontheleftsideofthemirror.$

Magnification is defined as:

$M=\frac{imageheight}{objectheight}=-\frac{v}{u}=-\frac{-37.5}{-25}\phantom{\rule{0ex}{0ex}}Objectheight=7cm\phantom{\rule{0ex}{0ex}}Hence,\phantom{\rule{0ex}{0ex}}imageheight=-\frac{-37.5}{-25}\times 7cm=-10.5cm\phantom{\rule{0ex}{0ex}}Negativesignimpliesthattheimageisinvertedwhichisthecasewhentheimageisreal.\phantom{\rule{0ex}{0ex}}Thus,\phantom{\rule{0ex}{0ex}}Theimageis10.5cminheightandreal.$

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