Question

An object 4 cm in height is placed at a distance of 16 cm from the concave mirror which produce a real image 5 cm find the position of the object and the focal length of the object

Answer 1

$\huge{\underline{\underline{\red{\mathcal{Solution}}}}}:-$

★ $Height \ of \ object \ \sf h_1$ = 4cm

★ $Height \ of \ image \ \sf h_2$ = 5cm

★ $Object \ distance, \ u = 16cm$

★ $Image \ distance = v$

Magnification = $\dfrac{ \sf h_1}{ \sf h_2}$ = $\dfrac{-v}{u}$

$\implies$ = $\dfrac{5}{-4}$ = $\dfrac{-v}{-16}$

• $-16 × 5 = -4 × -v$

• $-60 = 4v$

• $v = \cancel {\dfrac{-60}{4}}$

• $\large\bold{\underline{\boxed{\sf{\pink{v = -15cm}}}}}$

★ Using mirror formula :-

$\dfrac{1}{u} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{1}{-16} + \dfrac{1}{-15} = \dfrac{1}{f}$

$\dfrac{(15) + (16)}{-240} = \dfrac{1}{f}$

$\dfrac{31}{-240} = \dfrac{1}{f}$

$f = \dfrac{-240}{31}$

• $\large\bold{\underline{\boxed{\sf{\pink{f = -7.74cm}}}}}$

Hence,

$\large\bold{\underline{\sf{\red{Focal \; length \; is \; -7.74cm \; and \; image \; distance \; is \; 15cm.}}}}$

__________________________________________