Question

find the value of
$( x ^ { 3 } - \frac { 1 } { x ^ { 3 } } )$
if
$\frac { x } { 2 } = \frac { 1 } { 2 x } + 1$
​

Answer 1

Answer:

u=−30cm and magnification m=(−v/u)=−3 so v=(3u)

image is real so m should be negative v should be negative so v=−90cm

putting the above values of v and u in the Mirror equation,

f

1

=

u

1

+

v

1

we get focal length f=−22.5cm and image position v=−90cm

sorry I can't understand what you are saying