Question

A 20 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 25 cm . The distance of the object from the lens is 10 cm. Find the position , nature and size of the image forward .

Answer 1

Given :-

◉ $\quad$ Height of object, h = +20 cm

◉ $\quad$ Focal length of convex lens, f = +25 cm

◉ $\quad$Distance of object from the lens, u = -10 cm

To Find :-

◉$\quad$ Position, Nature and size of the image formed.

Solution :-

$\setlength{ \unitlength}{1cm} \thicklines \begin{picture}(10,10) \put(0,2){ \line(1,0){8}} \put(4,2){ \circle*{0.1}} \qbezier(4.1,4)(5,2)(4.1, 0) \qbezier(4.1,4)(3.2,2)(4.1,0) \put(4,4){ \line(0, - 1){4}} \put(2,1.5){ \sf f_{1} } \put(6,1.5){ \sf f_{2} } \put(2,2){ \circle*{0.1}} \put(6,2){ \circle*{0.1}} \put(2.5,2){ \vector(0,1){1}} \put(1.5,2.3){ \sf 20 cm} \put(2.5, 2.2){ \vector(1, 0){1.5}}\put(3,2.2){ \vector( - 1,0){0.5}} \put(2.8,2.3){ \sf 10 cm} \end{picture}$

We know,

$\qquad \boxed{\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} }$

$\sf \Rightarrow \quad \dfrac{1}{25} = \frac{1}{v} - \bigg( - \frac{1}{10} \bigg) \\ \\ \sf \Rightarrow \quad \frac{1}{25} = \frac{1}{v} + \frac{1}{10} \\ \\ \sf \Rightarrow \quad \frac{1}{v} = \frac{1}{25} - \frac{1}{10} \\ \\ \sf \Rightarrow \quad \frac{1}{v} = - \frac{3}{50}$

Hence, v = -50/3 cm or -16.67 cm

Properties of image :-

☛$\quad$ Nature :- Virtual and Erect

☛$\quad$ Size :- Magnified

☛$\quad$ Position :- 16.67 cm from the lens

Some Information :-

☛ $\quad$ Focal length of convex lens is positive while focal length of concave lens is negative.

☛$\quad$ If the object is between the focal point and pole then the image formed will be Magnified, virtual and erect and the position of image will be behind the object.