Question

Optics physics question!

84. ...A lens of focal length + 15.0 cm is 10.0 cm to the left of a second lens of focal length -15.0 cm. (a) Where is the final image of an object that is 30.0 cm to the left of the positive lens? (b) Is the image real or virtual? (c) How do the image?s size and orientation compare to those of the original object? Explain your answer. (d) How must you be oriented to see the image?

Answer 1

For the first lens,

Note that      
      
1/f = 1/do + 1/di      
      
where      
      
f = focal length =    15   cm
do = object distance =    3.00E+01   cm
di = image distance      
      
Thus, solving for di,      
      
di =    30   cm
      
Thus, the magnification, m = -di/do,      
      
m =    -1  

Thus, for the second lens,

Note that      
      
1/f = 1/do + 1/di      
      
where      
      
f = focal length =    -15   cm
do = object distance =    -2.00E+01   cm
di = image distance      
      
Thus, solving for di,      
      
di =    -60   cm

Thus, the final image is 60 cm to the right of the second lens.   [ANSWER, PART A]
      
As di < 0, it is VIRTUAL.   [ANSWER, PART B]

Thus, the magnification, m = -di/do,      
      
m =    -3  

Thus, the overall magnification is

mtot = m1 m2 = (-1)(-3) = 3

Thus, the final image is ERECT AND ENLARGED 3 times.   [ANSWER, PART C]

You must be upright facing to the right to see the image.   [ANSWER, PART D]