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a double convex lens made of glass of refractive index 1.6 has its both the surface  of equal radii of curvature of 30 cm and height of the object is 5 cm and place at the distancce of 12.5 cm .find the hight of the image

Asked by Tripurari 2nd February 2018, 9:36 PM
Answered by Expert
Answer:
Use lens maker formula to find focal length
begin mathsize 12px style 1 over f space equals space open parentheses mu minus 1 close parentheses open parentheses 1 over R subscript 1 minus fraction numerator begin display style 1 end style over denominator begin display style R subscript 2 end style end fraction close parentheses space equals space 0.6 cross times 2 over 30 semicolon space left parenthesis R subscript 1 space p o s i t i v e comma space R subscript 2 space n e g a t i v e space a s space p e r space c a r t e s i a n space s i g n space c o n v e n t i o n right parenthesis
f space equals space 25 space c m end style
 
begin mathsize 12px style 1 over f equals 1 over v minus 1 over u space semicolon
1 over 25 equals 1 over v plus fraction numerator 1 over denominator 12.5 end fraction semicolon space left parenthesis space u space i s space n e g a t i v e space a s space p e r space c a r t e s i a n space s i g n space c o n v e n t i o n right parenthesis
y o u space w i l l space g e t space v space equals space minus 25
end style
negative sign shows virtual image.
 
magnification = v/u = (-25) / (-12.5) = 2 ; hence height of the image 2×5 = 10 cm
Answered by Expert 5th February 2018, 11:10 AM
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