what is the image distance and image height if a 16.7 cm tall object is placed a distance of 39.5 cm from a diverging lens having a focal length of -9.25 cm?
Answer (1)
LENS[tex]••••••••••••••••••••••••••••••••••••••••••••••••••[/tex] Answer: Image Distance = -12.08 cm(Virtual Image) Image Height = 5.18 cm[tex]••••••••••••••••••••••••••••••••••••••••••••••••••[/tex] Solution: Find the image distance of the object that is placed from a diverging lens using the Lens Formula. [tex] \rm \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ [/tex]Substitute the given. [tex] \rm - \frac{1}{9.25} = - \frac{1}{39.5} + \frac{1}{v} \\ [/tex][tex] \rm - \frac{1}{9.25} + \frac{1}{39.5} = \frac{1}{v} \\ [/tex][tex] \rm - \frac{242}{2923} = \frac{1}{v} \\ [/tex][tex] \rm - 242v = 2923[/tex][tex] \rm \frac{ - 242v}{242} = \frac{2923}{ - 242} \\ [/tex][tex] \rm v = - 12.08 \: cm[/tex]Since the image distance of the object is negative, therefore the image formed at the same side of the object from the lens (Virtual Image). Find the image height of the object using the Magnification Formula. [tex] \rm m= - \frac{v}{u} \\ [/tex]Solve for the magnification first by substituting the given. [tex] \rm m= \frac{12.08}{39.5} \\ [/tex][tex] \rm m= 0.31[/tex]Find the image height of the object. [tex]\rm H_i = m \cdot H_o[/tex][tex]\rm H_i =0.31 \cdot 16.7[/tex][tex]\rm H_i =5.18 \: cm[/tex]Therefore, the image height of the object is 5.18 cm[tex]••••••••••••••••••••••••••••••••••••••••••••••••••[/tex] I HOPE IT HELPS :)
