Question

An object when placed in front of convex lens forms a real image of 0.5 magnification. If the distance of the image from the lens is 24 cm, Calculate Focal Length of Lens.​

Answer 1

$\text{u =?} \\ \text{ v = 24cm} \\ \text{ m = 0.5} \\ \text{ f = ?}$

$\implies \text{m} = \frac{ \text{v}}{ \text{u}} = \frac{24}{ \text{u}}$

$\implies 0.5 = \frac{1}{2} = \frac{24}{ \text{u}} \\ \\ \therefore \text{u} = 48 \text{cm}$

$\implies \frac{1}{ \text{f}} = \frac{1}{ \text{v}} - \frac{1}{ \text{u}}$

$\implies \text{f} = \frac{1}{24} - \frac{1}{48} = \frac{2 - 1}{48} = \frac{1}{48}$

$\text{f} = 48 \text{cm}$

$\underline{ \text {According To Question,}}$

$\text{24cm is the distance of object from lens,}$

$\text{m} = \frac{ \text{v}}{ \text{u}} \\ \\ \implies \frac{1}{2} = \frac{ \text{u}}{24}$

$\therefore \text{v} = \frac{24}{2} = 12 \text{cm}$

$\implies \frac{1}{ \text{f}} = \frac{1}{ \text{v}} - \frac{1}{ \text{u}}$

$\implies \frac{1}{ \text{f}} = \frac{1}{12} - ( \frac{1}{ - 24} )$

$\implies \frac{1}{ \text{f}} = \frac{1}{12} + \frac{1}{24} = \frac{2 + 1}{24} = \frac{3}{24}$

$\implies \text{f} = \frac{24}{3} = 8 \text{cm}$

Hope this helps