Question

an object is placed perpendicular to the principal axis of a convex lens of foacl length 8 cm. the distence of the object from lens is 12 cm find position and nature of image... Guys plss hlp ​

Answer 1

Given :

• Focal length of convex lens is 8 cm.
• The distance of the object from lens, u = -12 cm.

To Find :

• Position of image.
• Nature of the image.

Solution :

Using the lens formula,

$\large{\boxed{\sf{\dfrac{1}{f}\:=\:\dfrac{1}{v}-\dfrac{1}{u}}}}$

Block in the values,

$\longrightarrow$ $\sf{\dfrac{1}{8}=\dfrac{1}{v}-\dfrac{1}{-12}}$

$\longrightarrow$ $\sf{\dfrac{1}{8}\:=\:\dfrac{-12-v}{-12v}}$

$\longrightarrow$ $\sf{-12v=8(-12-v) }$

$\longrightarrow$ $\sf{-12v=-96-8v}$

$\longrightarrow$ $\sf{-12v+8v=-96}$

$\longrightarrow$ $\sf{-4v=-96}$

$\longrightarrow$ $\sf{4v=96}$

$\longrightarrow$ $\sf{v=\dfrac{96}{4}}$

$\longrightarrow$ $\sf{v=24}$

$\large{\boxed{\sf{Position\:of\:image\:=\:24\:cm\:on\:+X\:of\:optical\:centre\:}}}$

$\large{\boxed{\sf{Nature\:of\:image\:is\:real\:and\:inverted}}}$

Answer 2

Explanation:

Given:

✍️Focal length of convex lens (f) is 8 cm.

✍️ Distance of the object from lens (u) is 12cm .

Formula used:

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

To find:

position and nature of image.

Solution:

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

$\frac{1}{8} = \frac{1}{v} - \frac{1}{ - 12}$

$\frac{1}{8} = \frac{ - 12 - v}{ - 12v}$

$- 12v = - 96 - 8v$

$- 12v + 8v = 96$

$4v = 96$

$v = \frac{96}{4}$

$v = 24$

✴️The positive sign of v shows that the images is formed at a distance of 24 cm on the other side of optical centre.

✴️The images is real and inverted.