Question

7. An object is located 14.5 cm in front of a convex mirror, the image being 6.55 cm behind the mirror. A second object, twice as tall as the first one, is placed in front of the mirror, but at a different location. The image of this second object has the same height as the other image. How far in front of the mirror is the second object located? X43.69462 cm

Answer 1

We know that magnification due to the mirror Here v= 6.55em, up-14.5cm mi= b = -6.55 -0.452 - 14.5 = 6.452 for and object me b e According to the - question --> from ea, v 0.452 = -4 = =-0. 2264 - 3 Now Using mirror equation 1 1 f - V 1 u 1 6.55 - 1 = 14.5-6.55 14.5 6.55x14.5 = 7.95 6.55X145 of ) t = lligsch - (4) Argain for sento and Object It. & " f . - 6.2264 ! " (using eq,0) -> mis lupust") (ling f=1.5cm] 11.95 0.226 -)).9 cm S = - 40.93cm 1421= 40.93 cm Here u-object location - image location f- focus