A converging lens from a real and inverted image of an object at a distance of 100cm. Where should an object be placed in front of the lens so that the size of the image is twice the size of object. Draw the ray diagram and also calculate the power of a lens
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Let h2 be the height of image and h1 be the hight of object
as condition h2/h1=2
As magnification m=h2/h1=v/u
v/u=2,,v=2u
as v=100 cm,,as object placed at right side then u always negative
u=-50cm
by lens formula
1/f=1/v-1/u
f=100/3
power =100/f=3D
object should place at 50cm from pole of lens
Sorry for ray diagram,It is typical to draw on this page
as condition h2/h1=2
As magnification m=h2/h1=v/u
v/u=2,,v=2u
as v=100 cm,,as object placed at right side then u always negative
u=-50cm
by lens formula
1/f=1/v-1/u
f=100/3
power =100/f=3D
object should place at 50cm from pole of lens
Sorry for ray diagram,It is typical to draw on this page
kaushikravikant:
is there any need of improvement in this answer
could u tell where the objct should b placed whether on F or P or C.....
how did u get 50
as mention v=2u, we have given v=100cm then simply we find u
as f=33.3 then c=2f=66.6 as u=50cm.From all this we conclude that object is placed between f and C
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