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190. A 5.00 cm object is placed 25.0 cm in front of a convex lens with a focal length of 10.0 cm.  The image formed is
    
A. 3.34 cm tall, 16.7 cm from lens, real, and inverted. 1        1        1
---- + ---- = ----
do      di        f

    
The focal length is positive for converging lens (convex) and mirrors (concave).  It is negative for diverging lens and mirrors.

The object distance is positive and we are solving for di

di = ( f -1  -  do -1 ) -1 =  

di = ( (10.0 cm) -1 - (25.0 cm) -1 ) -1 

di = 16.7 cm   since it is positive, it is a real image on the other side of the lens.

hi          di
---- = - ----- 
ho         do 

    
          di h          16.7 cm x 5.00 cm
hi = - -------- =  - --------------------------
            d                   25.0 cm

    
hi  =  -  3.34 cm

The negative sign shows the the image will be inverted. 

B. 3.34 cm tall, 16.7 cm from lens, virtual, and upright.
C. 1.43 cm tall, 7.14 cm from lens, real, and inverted.
D. 1.43 cm tall, 7.14 cm from lens, virtual, and upright.