A concave lens has focal length of 20cm. At what distance from the lens a 5cm tall object be placed so early hat it forms an image at 15cm the images formed?
Answers
Answer:
f=-20cm
v=-15
1 over v minus 1 over u equals 1 over f
fraction numerator begin display style 1 end style over denominator begin display style negative 15 end style end fraction minus fraction numerator begin display style 1 end style over denominator begin display style u end style end fraction equals fraction numerator begin display style 1 end style over denominator begin display style negative 20 end style end fraction
fraction numerator begin display style 1 end style over denominator begin display style u end style end fraction equals fraction numerator begin display style 1 end style over denominator begin display style 20 end style end fraction minus fraction numerator begin display style 1 end style over denominator begin display style 15 end style end fraction
fraction numerator begin display style 1 end style over denominator begin display style u end style end fraction equals fraction numerator begin display style 3 minus 4 end style over denominator begin display style 60 end style end fraction
fraction numerator begin display style 1 end style over denominator begin display style u end style end fraction equals fraction numerator begin display style negative 1 end style over denominator begin display style 60 end style end fraction
u equals negative 60
m equals v over u equals fraction numerator H e i g h t space o f space i m a g e left parenthesis h subscript i right parenthesis over denominator H e i g h t space o f space o b j e c t left parenthesis h subscript o right parenthesis end fraction
fraction numerator negative 15 over denominator negative 60 end fraction equals h subscript i over 5
h subscript i equals 5 over 4 equals 1.25 c m.
was that help ful
1/f = 1/v-1/u
1/(-20) - 1/-15= - 1/u
(15-20)/300 = - 1/u
1/u = - 5/300
u = - 60 cm
hi/ho= v/u
hi/5 = -15/-60
hi=5/4 cm
=1.25 cm ☑️☑️