When an object is placed 40 cm away from a spherical mirror an image of magnification $\frac{1}{2}$ is produced. To obtain an image with magnification of $\frac{1}{3}$, the object is to be moved :

<p>To solve this problem, start by using the magnification formula for spherical mirrors:</p> <p>$ m = \frac{f}{f - u} $</p> <p>Given that the initial magnification $ m = \frac{1}{2} $, and the object distance $ u = -40 $ cm (the negative sign indicates distance measured against the incoming light), we can write:</p> <p>$ \frac{1}{2} = \frac{f}{f - (-40)} $</p> <p>This simplifies to:</p> <p>$ \frac{1}{2} = \frac{f}{f + 40} $</p> <p>Cross-multiplying gives:</p> <p>$ f + 40 = 2f $</p> <p>This leads to:</p> <p>$ f = 40 \text{ cm} $</p> <p>To find the new object distance for a magnification of $ m = \frac{1}{3} $, we use the same formula:</p> <p>$ m = \frac{f}{f - u} $</p> <p>Substitute the known values:</p> <p>$ \frac{1}{3} = \frac{40}{40 - u} $</p> <p>Cross-multiplying gives:</p> <p>$ 40 - u = 120 $</p> <p>Solving for $ u $ results in:</p> <p>$ u = -80 \text{ cm} $</p> <p>Thus, the object needs to be placed 80 cm from the mirror, opposite to the initial placement of 40 cm. To achieve this repositioning, the object needs to be moved 40 cm further from its initial 40 cm distance, totaling a move of 40 cm away from the mirror.</p>