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CBSE Class 12-science Answered

Dera sir, please explain the displacement method to find the focal length of the convex lens in detail
Asked by modi72879 | 22 Feb, 2018, 10:03: PM
answered-by-expert Expert Answer
Let us fix the position of object and place the screen to get the enlarged image first. Also let us fix the position of screen where we get the enlarged image.
Let D be the distance bewteen object and screen. Let us mark the posistion of lens x1. Then let us move the lens away from the object to get a diminshed image. Let this position of lens be x2.  let X be the distance between the lens positions x1 and x2.
Let v be the distance between image and lens. Let u be the distanve between object and lens

begin mathsize 12px style 1 over v plus 1 over u equals 1 over f l e t space u s space e l i m i n a t e space v space b y space s u b s t i t u t i n g space v space equals space D minus u fraction numerator begin display style 1 end style over denominator begin display style D minus u end style end fraction plus 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 f end style end fraction space semicolon space w e space g e t space t h e space e q u a t i o n space space space u squared minus D u plus f D space equals 0 t h e space s o l u t i o n space f o r space t h e space a b o v e space q u a d r a t i c space e q u a t i o n space i s space comma space space u space equals space fraction numerator D plus-or-minus square root of D squared minus 4 f D end root over denominator 2 end fraction w h e n space D space equals space 4 f comma space w e space g e t space o n l y space o n e space p o s i t i o n space o f space l e n s space t o space g e t space i m a g e. space T h i s space c o r r e s p o n d s space t o space p l a c i n g space t h e space o b j e c t space a t space 2 f space a n d space g e t t i n g space t h e space i m a g e space a t space 2 f space o n space t h e space o t h e r space s i d e. H e n c e space f o r space d i s p l a c e m e n t space m e t h o d space w e space n e e d space D greater than 4 f. space w h e n space t h i s space c o n d i t i o n space i s space s a t i s f i e d space w e space g e t u 1 space equals fraction numerator D minus square root of D squared minus 4 f D end root over denominator 2 end fraction space semicolon space c o r r e s p o n d i n g space v 1 space space equals space D minus space u 1 space equals fraction numerator begin display style D plus square root of D squared minus 4 f D end root end style over denominator begin display style 2 end style end fraction a f t e r space c h a n g i n g space t h e space l o c a t i o n u 2 space equals fraction numerator begin display style D plus square root of D squared minus 4 f D end root end style over denominator begin display style 2 end style end fraction space semicolon space c o r r e s p o n d i n g space v 2 space space equals space D minus space u 2 space equals fraction numerator begin display style D minus square root of D squared minus 4 f D end root end style over denominator begin display style 2 end style end fraction n o w space t h e space d i s p l a c e m e n t space space space space X space equals space v 1 minus u 1 space equals u 2 minus v 2 space space equals space square root of D squared minus 4 f D end root semicolon h e n c e space w e space g e t space f o c a l space l e n g t h comma space space f space equals space fraction numerator D squared minus X squared over denominator 4 D end fraction end style
Hence from the known values of D and x we can determine the focal length of convex lens
Answered by Thiyagarajan K | 23 Feb, 2018, 07:42: AM

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