Question

an object is placed at a distance of 30 cm from the lens its image is formed at a distance of 10 cm on the same side of the lens find the focal length of the lens used​

Answer 1

★ Figure refer to attachment

GIVEn

An object is placed at a distance of 30 cm from the lens its image is formed at a distance of 10 cm on the same side of the lens.

TO FIND

Find the focal length and name the lens used

SOLUTIOn

(★) In convex lens, when object kept between focus and optical centre then image will formed on same side of lens as the object

Nature of image

• Nature = Virtual and erect
• Size = Enlarge
• Position = Same side of lens as the object

Now,

• Object distance (u) = - 30cm
• Image distance (v) = + 10cm
• Focal length (f) = ?

Apply lens formula

$\implies\sf \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{1}{10}-\dfrac{(-1)}{30}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{1}{10}+\dfrac{1}{30}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{3+1}{30}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{1}{f}=\cancel\dfrac{4}{30} \\ \\ \\ \implies\sf \dfrac{1}{f}=\dfrac{2}{15} \\ \\ \\ \therefore\sf \:f = +7.5cm$

Hence, the focal length of convex lens is

+ 7.5cm

Answer 2

$\large\red{\underline{{\boxed{\textbf{Brainliest\:Answer}}}}}$

$\rule{200}{2}$

$\red{\bold{\underline{\underline{ \: Details \: Given\::}}}}$

An Object is placed at a distance of 30 cm from the lens its image is formed at a distance of 10 cm on the same side of the lens

$\red{\bold{\underline{\underline{Nature \:of \:Image \::}}}}$

• Nature = Virtual and erect
• Size = Enlarge
• Position = Same side of lens as the object

$\red{\bold{\underline{\underline{Required \: Solution\::}}}}$

In a convex lens when object kept between focus and optical centre then image will formed on same side of lens as the object

Therefore

Object distance (u) = - 30cm

Image distance (v) = + 10cm

Focal length (f) = ?

$\rule{200}{2}$

$\red{\bold{\underline{\underline{Using \: Lens \: Formula \::}}}}$

$\\ \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} \\ \\ \dfrac{1}{10}-\dfrac{(-1)}{30}=\dfrac{1}{f} \\ \\ \\ \\ \dfrac{1}{10}+\dfrac{1}{30}=\dfrac{1}{f} \\ \\ \\ \\ \dfrac{3+1}{30}=\dfrac{1}{f} \\ \\ \\ \\ \dfrac{1}{f}=\cancel\dfrac{4}{30} \\ \\ \\ \\ \dfrac{1}{f}=\dfrac{2}{15} \\ \\ \\ \therefore\ \:f = +7.5cm$

$\rule{200}{2}$

$\boxed{\bold{\red{Keep\: Asking\: - \: Be \: Brainly}}}$