Question

an object is placed at 5cm at a distance of 30 from a converging lens of focal length 10cm find the postion and nature

who can solve this as a numerical way​

Answer 1

Solution :

$\underline{\bf{Given\::}}}$

• Size of object, (h1) = 5 cm
• Distance of object from lens, (u) = 30 cm
• Focal length, (f) = 10 cm

$\underline{\bf{Explanation\::}}}$

Using formula of the lens :

$\boxed{\bf{\frac{1}{f} =\frac{1}{v} -\frac{1}{u} }}}$

$\longrightarrow\sf{\dfrac{1}{10} =\dfrac{1}{v} -\dfrac{1}{-30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{10} -\dfrac{1}{30} }\\\\\\\longrightarrow\sf{{\dfrac{1}{v} =\dfrac{3-1}{30} }}\\\\\\\longrightarrow\sf{{\dfrac{1}{v} =\dfrac{2}{30} }}\\\\\longrightarrow\sf{2v=30}\\\\\longrightarrow\sf{v=\cancel{30/2}}\\\\\longrightarrow\bf{v=15\:cm}$

∴ Image is 15 cm far from the lens on other side,  i.e, behind the lens image is real & inverted.

Now;

Using formula of the linear magnification :

$\mapsto\sf{m=\dfrac{Height\:of\:image\:(I)}{Height\:of\:object\:(O)} =\dfrac{Distance\:of\:image}{Distance\:of\:object} =\dfrac{v}{u} }$

$\mapsto\sf{\dfrac{h_2}{h_1} =\dfrac{-v}{u} }\\\\\mapsto\sf{\dfrac{h_2}{5} =\dfrac{15}{30} }\\\\\mapsto\sf{30h_2=15\times 5}\\\\\mapsto\sf{30h_2=75}\\\\\mapsto\sf{h_2=\cancel{75/30}}\\\\\mapsto\bf{h_2=2.5\:cm}$

∴ Height of image is 2.5 cm, the image is real & inverted .