SIGN CONVENTION Distance of object from lens is

SIGN CONVENTION Distance of object from lens is U Distance of image from lens is V Distance of second focus is F

AN OBJECT OF HEIGHT 4. 0 CM IS PLACED AT A DISTANCE OF 24 CM IN FRONT OF A CONVEX LENS OF FOCAL LENGTH 8 CM. FIND (a) POSITION AND SIZE OF THE IMAGE and (b) THE CHARACTERISTICS OF THE IMAGE 8 24 WHAT IS GIVEN? ? • O = + 4 0 cm • F = + 8 cm • U = - 24 cm SUBSTITUTE THESE VALUES IN THE FORMULA AND FIND OUT!!

AT WHAT POSITION A CANDLE OF LENGTH 3 CM BE PLACED IN FRONT OF A CONVEX LENS SO THAT ITS IMAGE OF LENGTH 6 CM BE OBTAINED ON A SCREEN PLACED AT A DISTANCE 30 CM BEHIND THE LENS? WHAT IS THE FOCAL LENGTH OF THE LENS? 3 6 30 SO WHAT IS GIVEN? O – 3 cm I – 6 cm V = 30 cm We know magnification m = I = V O U SO V /U = 6/3 = 2 cm; since V = 30 , U = 15 cm We know 1/v – 1/u = 1/f; now find f;

POWER OF A LENS WE NORMALLY SAY THE POWER OF A LENS IS =+ 2. 25; POWER OF LENS = - 1. 25; WHAT DOES IT MEAN?

WHEN A LIGHT PASSES THROUGH A LENS THERE IS A DEVIATION. CAN WE SAY MORE THE DEVIATION MORE THE POWER? IF THE LENS IS VERY THICK, WHAT ABOUT THE FOCAL LENGTH? IF THE LENS IS VERY THIN WHAT ABOUT THE FOCAL LENGTH? THINK!!

POWER OF A LENS POWER OF LENS (in D) = 1 FOCAL LENGTH ( in METRES) POWER IS POSITIVE – DEVIATED TOWARDS THE CENTRE – CONVEX LENS POWER IS NEGATIVE – DEVIATED AWAY FROM THE CENTRE - CONCAVE LENS