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Q.
If in the following figure, height of object is $H_1 = +2.5cm$, then height of image $H_2$ formed is
Ray Optics and Optical Instruments
Solution:
By concave mirror formula
$ \frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
Here $ u = - 1.5 f, f = - f$
we have $ \frac{1}{(-f)}=\frac{1}{v}+\frac{1}{(-1.5 f)}$
$\therefore \frac{1}{v}=\frac{1}{3f} $ or $v=-3f$
Now, magnification in mirror is given by
$ m= \frac{I}{O}=\frac{v}{-u}$
where I is size of image and O is size of object.
Here $ 0 = + 2.5\, cm$
$\therefore \frac{I}{2.5}=\frac{-3 f}{(-1.5f)}$ or $I=-5\, cm$
