Physics, asked by Anonymous, 1 year ago

Object of height 4 cm is placed at a distance of 30 cm from the optical centre of a convex lens of focal length 20cm draw a ray diagram to find the position and size of the image formed mark optical Centre O and principal focus f on a diagram also find the approximate ratio of size of the image to size of the object

Answers

Answered by kavya110
46
u = -30 cm
f = 20cm
h= 4 cm.

1/v- 1/u = 1/f
1/v= 1/20-1/30
= (3-2)/ 60
1/v = 1/60
v= 60 cm.

m=v/u = h'/h
60/-30= h' / 4
-2 = h'/4
h' = -8

ratio = - 2:1

Anonymous: Thank god..my correct
kavya110: i did 2:1
Anonymous: ya..me too ratio cannot be negative
Answered by Anonymous
56

\bf\huge\boxed{\boxed{\bf\huge\:See\:Attached\:Image}}}



\bf\huge\frac{1}{v} - \frac{1}{u} = \frac{1}{f}



\bf\huge\frac{1}{v} = \frac{1}{f} + \frac{1}{u}



\bf\huge\frac{1}{v} = \frac{1}{20} - \frac{1}{30}



 \bf\huge\frac{1}{v} = \frac{3-2}{60}



 \bf\huge\frac{1}{v} = \frac{1}{60}



 \bf\huge\frac{h1}{h} = \frac{v}{u}



 \bf\huge\frac{h1}{4} = \frac{60}{-30}



\bf\huge\texttt h1 = -8cm



 \bf\huge\texttt Ratio = \frac{Height\:of\:Image}{Height\:of\:Object}



 \bf\huge\frac{-8}{4} = \frac{-2}{1}



 \bf\hugeRatio = -2 : 1



 \bf\huge m = -2 : 1



\bf\huge\boxed{\boxed{\bf\huge\:Image=\:Real\:Inverted\:and\:Enlarged}}}



Attachments:
Similar questions