Question

An object is
imagned by a lens on
screen placed 12 cm from the lens when
the lers is
moved
2cm
be moved
object, the
screen must
closer
away from The
2 cm
( to the
Object
to reform it
find the focal length of the rene
c) 8cm
D) 6cm
B) SCM
ист​

Answer 1

Answer:

imagned by a lens on

screen placed 12 cm from the lens when

the lers is

moved

2cm

be moved

object, the

screen must

closer

away from The

2 cm

( to the

Object

to reform it

Answer 2

Answer:

        The focal length of the lens is 4cm.

Explanation:

       The lens formula is $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$.

     Now, the new lens

                                 $u'=u+12\\ v'=12-(2+8)\\=8$

     So,

            $\frac{1}{8}+\frac{1}{u+12}=\frac{1}{12}+\frac{1}{u} \\\frac{u+10}{8(u+2} =\frac{u+12}{12u} \\12u^{2} +120u=8[u^{2} +14u+24]\\4u^{2} +8u-192=0\\u^{2} +2u-48=0\\u=6$

    The equation is

                         $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\=\frac{1}{6} +\frac{1}{12} \\=\frac{3}{12}\\ f=4$