Question

Question 2, chap 136, sect 8. part 1 of 1 10 points The distance between an object and its up- right image is 12.2 cm. If the magnification is 0.427, what is the focal length of the lens being used to form the image? Answer in units of cm.

Answer 1

Solution - distance between an object and its upright linage is (d) = 12.2cm Magnification (m) = 0.427 If image formed is upright and small sige. then this lens in (diverging), object concave 7 Image F 1 0 Here (u-v=d) ku object distance au Image distance au As , magnification (m) = height of Image Image ditame height of object object olistance m= = 0.427 и v=0.427 u 12.2 From leg 0 and ② u-v=d 1- [16.427)u (0.573) u object distance (U)= 12.2 Image distance (0) = und = 12:2 - 12.2 21.3 cm 0.573 21.3-12.2 = 9.1 an.

By 1 1 Both object and Image are in left side of lens : So u=- 21.3cm v= -a-l cum using lens formula :- Lud f v -9.1 - (21.3) / $ f 9.1 21.3 fa 21.3 X 9.) - 15.9 cm 7-21.379.1) и - + Focal length (f) = -15.9 cm