Question

An object of height 4.0 cm is placed at a distance of 30 cm from the optical centre

‘O’ of a convex lens of focal length 20 cm. Draw a ray diagram to find the position

and size of the image formed. Mark optical centre ‘O’ and principal focus ‘F’ on the

diagram. Also find the approximate ratio of size of the image to the size of the object.

Answer 1

the height of the image is 1.6cm

Answer 2

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

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

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

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

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

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

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

$\bf\huge\texttt h1 = -8cm$

$\bf\huge\textttRatio = \frac{Height\:of\:Image}{Height\:of\:Object}$

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

$\bf\hugeRatio = -2 : 1$

$\bf\hugem = -2 : 1$

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