Question

A concave lens has focal length of 120m. At what ustalice should the object
from the lens be placed so that it
forms an image at 8cm from the lens? Also,
find the magnification produced by the lens,​

Answer 1

Answer:

$\frac{16}{15}$

Explanation:

Given :

• Focal length of the concave lens = - 120 cm (- because it is concave lens) = f (120m is a very large value so 120 cm must be considered as you have done a typo error)
• Distance of image from the lens = 8 cm = v

To find :

• Magnification of the object

Using lens formula which says:

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

Substituting the values :

$\frac{1}{8} - \frac{1}{u} = \frac{1}{-120}$

$- \frac{1}{u} = \frac{-1}{120} - \frac{1}{8}$

$- \frac{1}{u} = \frac{-1}{120} - \frac{15}{120}$

$- \frac{1}{u} = \frac{ - 16}{120}$

Cross multiplying :

-120 = - 16u

$u = \frac{120}{16}$

Now we have to find Magnification :

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

$m = \frac{8}{ \frac{ 120}{16} }$

$m = \frac{ 8}{120} \times {16}$

$m = \frac{16}{15}$

The magnification is 16/15

Answer 2

Given :-

• Focal length (f)= 120 cm ( You had written it's "m" by mistake )

• Image distance ( v) = 8cm

To Find :-

• Object distance

• Magnification

★ Find object distance (u)

According to lens formula -

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

$\longmapsto\sf \dfrac{-1}{120}= \dfrac{1}{8}-\dfrac{1}{u}\\ \\ \longmapsto\sf \dfrac{1}{u}= \dfrac{1}{8}+\dfrac{1}{120}\\ \\ \longmapsto\sf \dfrac{1}{u}= \dfrac{15+1}{120}\\ \\ \longmapsto\sf \dfrac{1}{u}= \dfrac{16}{120}\\ \\ \longmapsto\sf u= \cancel{\dfrac{120}{16}}= \dfrac{60}{8}\\ \\ \longmapsto\sf u = \dfrac{15}{2} cm \ or \ 7.5 cm$

$\boxed{\sf\ Object\ distance = \dfrac{15}{2}cm \ or \ 7.5 cm }$

Now find the magnification :-

$\boxed{\sf m= \dfrac{(v) \ Image \ distance }{(u)\ object \ distance } }$

$\longmapsto\sf m= \dfrac{v}{u}\\ \\ \bullet\sf v= 8cm \ ; \ \bullet\ u= \dfrac{15}{2}cm\\ \\ \longmapsto\sf m = \dfrac{8}{\dfrac{15}{2}}\\ \\ \longmapsto\sf m= 8\times \dfrac{2}{15}\\ \\ \longmapsto\sf m= \dfrac{8\times 2}{15}\\ \\ \longmapsto\sf m = \dfrac{16}{15}$

$\boxed{\sf \ Magnification= \dfrac{16}{15}}$