Solution a 1f =1v -1u1v =1f +1u =14 +1(-4 5)=14 -1(4 5)v=4 5×44 5-4 =36 cmb Let fB is focal length of diverging lens1(-fB) =1v' -1u'u' = v'fBv'+fB =8×28+2=1 6 cmSo distance between lens A and Bl = v-u'= 34 4 cm

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Solution

a . \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{4} + \frac{1}{(- 4.5)} = \frac{1}{4} - \frac{1}{(4.5)} v = \frac{4.5 \times 4}{4.5 - 4} = 36 c m b . L e t f_{B} i s f o c a l l e n g t h o f d i v e r g i n g l e n s \frac{1}{(- f_{B})} = \frac{1}{v'} - \frac{1}{u'} u' = \frac{v' f_{B}}{v' + f_{B}} = \frac{8 \times 2}{8 + 2} = 1.6 c m S o d i s \tan c e b e t w e e n l e n s A a n d B l = v - u' = 34.4 c m

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