Question

A converging lens has a focal length of 6.0 cm

A.) Locate the position of the images for the object distance of 4.0cm

B.) Find the height of the image of an object 1.0 cm high

C.) State whether the image is real, virtual, upright, or inverted

D.) Draw a detailed and proportional ray diagram for this situation

Answer 1

Givens Conver ging leng => Conver lens focal Length (f)=+6.0cm Object distance (U)=-4 cm Image distance (v) = ? Size of object (0) = 1 cm as locate the position of image- Let, Distance of image from lens = u We know that, - el- u le f & * tt = + (4) et at 11 2 -3 12 1 = -12 cm ANSWER (b) find height of image (I) - Height of object (O) = 1cm. We know that. magnification- 313 m = u HO Hla 1 (-12) (-4) - 3 cm Erect image ANSWER I

ce Us - 4 cm de v=-12cm (C) Nature of image Since, u is negative and o is also negative It means both are on same side thats why Image will be "Virtucel". ANSWER Since, Size of image (I) is calculafeol positive le both object and Image are erect je "upright" ANSWER I. 3cm It means (D) I 25 C F' 2f u. I → Image (Virtual and upright) O Object u Object distance from bens =-4cm Image distance from lens = -12cm

In the ray diagram, light rays actually not meet at right side but pretend to meet on back side ( left side of lens) at 2F ( v = -12cm).

Positive and negative sign are taken using sign convention.