Question

Q.4 An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image. Draw ray diagram also.
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Answer 1

Given data:

           u = -10cm   ( according to cartesian sign )

            f = -15cm

            v = ?

solution:

       from the mirror formula

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

by inter changing the formula

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

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

$\frac{1}{v} =\frac{1}{-15} -\frac{1}{-10} \\\\ \frac{1}{v} =-\frac{1}{15} +\frac{1}{10} \\\\\frac{1}{v} =-\frac{1}{15} +\frac{1}{10} \\\\\frac{1}{v} =-\frac{1X(2)}{15X(2)} +\frac{1X(3)}{10X(3)} \\\\\frac{1}{v} =-\frac{2}{30} +\frac{3}{30} \\\\\\\frac{1}{v} =\frac{-2+3}{30} \\\\\frac{1}{v} =\frac{1}{30}\\\\v=30 cm$

∴image formed at 30 cm

image formed by convex mirror always virtual and errect