Question

A convex lens
of focal length= 10cm forms an image
which is
four times the
size of object
what is the distance of object from lense​

Answer 1

Answer:

-12.5 cm

Explanation:

Given :

• Focal length = 10 cm
• Magnification = - 4 (lmage is real if magnification is negative)

To find :

• Distance of object from lens

$m = \frac{v}{u}$

V is known as image distance and u is object distance

$- 4 = \frac{ v}{u}$

V = - 4u

Lens formula says that :

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

$v = - 4u$

$\frac{1}{-4u} - \frac{1}{u} = \frac{1}{10}$

$\frac{1}{-4u} - \frac{-4}{-4u} = \frac{1}{10}$

$\frac{ 5}{-4u} = \frac{1}{10}$

By cross multiplying :

-4u = 50

$u = \frac{ 50}{-4}$

$u = -12.5 \: cm$

The distance of object from lens is equal to -12.5 cm

Answer 2

S O L U T I O N :

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

• A convex lens of focal length (f) = 10 cm
• Magnification = 4

$\bf{\large{\underline{\bf{To\:find\::}}}}}$

The distance of object from the lens (u).

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

When the image is real and erect from a convex lens, the magnification will be negative.

$\boxed{\bf{m=\frac{Distance\:of\:image\:(v)}{Distance\:of\:object\:(u)} }}}$

$\longrightarrow\sf{m=\dfrac{v}{u} }\\\\\longrightarrow\sf{-4=\dfrac{v}{u} }\\\\\longrightarrow\sf{v=-4u}$

We know that formula of the lens :

$\longrightarrow\sf{\dfrac{1}{f} =\dfrac{1}{v} -\dfrac{1}{u} }\\\\\\\longrightarrow\sf{\dfrac{1}{10} =\dfrac{1}{-4u} -\dfrac{1}{u} }\\\\\\\longrightarrow\sf{\dfrac{1}{10} =\dfrac{-1-4}{4u} }\\\\\\\longrightarrow\sf{\dfrac{1}{10} =-\dfrac{5}{4u} }\\\\\\\longrightarrow\sf{4u=-50}\\\\\\\longrightarrow\sf{u=\cancel{\dfrac{-50}{4} }}\\\\\\\longrightarrow\bf{u=-12.5}$

So;

The distance of the object will be 12.5 cm .