An object 1 m tall is placed on the principal axis of a convex lens and its 40 cm tall image is formed on the screen placed at a distance of 70 cm from the object. The focal length of the lens.

An object

1 m

tall is placed on the principal axis of a convex lens and its

40 cm

tall image is formed on the screen placed at a distance of

70 cm

from the object. The Focal length of the lens,

f = + 14.3 cm

Given,
Height of the object,

h = 1 m

Height of the real image formed on the screen,

h^{'} = - 40 cm

Distance from object to screen,

|u + v| = 70 cm

⇒

m = \frac{h^{'}}{h} = \frac{v}{u}

⇒

\frac{v}{u} = \frac{- 40 cm}{1 m}

= \frac{- 40 cm}{100 cm}

= - 0.4

⇒

v = - 0.4 u

|v| = |0.4 u|

⇒

|u + v| = |u + 0.4 u|

= |1.4 u|

= 70 cm

⇒

|u| = \frac{70}{1.4}

= 50 cm

And hence,

|v| = 70 - |u|

= 70 - 50

= 20 cm

As per sign convention,

u

- ve

and

v

+ ve

, i.e,

u = - 50 cm

and

v = + 20 cm

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

= \frac{1}{+ 20} - \frac{1}{- 50}

= \frac{1}{20} + \frac{1}{50}

= \frac{5 + 2}{100}

\Rightarrow \frac{1}{f} = \frac{7}{100}

f = \frac{100}{7}

= + 14.3 cm

An object $1 m$ tall is placed on the principal axis of a convex lens and its $40 cm$ tall image is formed on the screen placed at a distance of $70 cm$ from the object. The focal length of the lens.