object is placed in front of a diverging lens with a focal length of 6 CM. Find the image distance for an object distance of 9 cm. Answer in units of cm.
Solution:-
given
u = - 9 cm
f = -6 cm
lens formula
1v−1u=1f\frac{1}{v} - \frac{1}{u}=\frac{1}{f}v1−u1=f1
1v=1u+1f\frac{1}{v} =\frac{1}{u}+\frac{1}{f}v1=u1+f1
Putting values
1v=1−9+1−6\frac{1}{v} =\frac{1}{-9}+\frac{1}{-6}v1=−91+−61
1v=−2−318\frac{1}{v} =\frac{-2-3}{18}v1=18−2−3
V=−185cm\boxed{V = {-18\over 5}cm}V=5−18cm i.e image distance
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