Question

an object is placed at a distance of 60 cm from a concave lens of focal length 30 cm number 1 distance of image number 2 find nature position size​

Answer 1

$\sf{Given}$

$\rightarrow$ Object distance (u) = - 60 cm

$\rightarrow$ Focal length (f) = - 30 cm

$\sf{Find}$

• The distance of the image (v)
• Nature of position size

$\sf{Solution}$

An object is placed at a distance of 60 cm from a concave lens of focal length 30 cm.

In the above question, we were talking about the lens. And lens formula is : $\underline{\boxed{\rm{\frac{1}{f}\:=\:\frac{1}{v}\:-\:\frac{1}{u}}}}$

Here -

• f = focal length
• v = image distance
• u = object distance from the lens

Substitute the known values above to find the image distance from the lens.

$\implies\:\sf{\dfrac{1}{-30}\:=\:\dfrac{1}{v}\:-\:\dfrac{1}{-60}}$

$\implies\:\sf{\dfrac{-1}{30}\:=\:\dfrac{1}{v}\:+\:\dfrac{1}{60}}$

$\implies\:\sf{\dfrac{-1}{30}\:-\:\dfrac{1}{60}\:=\:\dfrac{1}{v}}$

$\implies\:\sf{\dfrac{-2\:-1}{60}\:=\:\dfrac{1}{v}}$

$\implies\:\sf{\dfrac{-3}{60}\:=\:\dfrac{1}{v}}$

$\implies\:\sf{\dfrac{-1}{20}\:=\:\dfrac{1}{v}}$

$\implies\:\sf{v\:=\:-20}$

As the image distance is negative. So, the image is virtual and erect.

•°• Image is located 20 cm away from the concave lens.

Magnification :

It is defined as the ratio of the height of the image with the height of the object.

i.e.

$\underline{\boxed{\rm{m\:=\:\frac{v}{u} \: = \: \frac{ h_{i}}{ h_{o} } }}}$

Substitute the known values in above formula

$\implies\:\sf{\dfrac{-20}{-60}}$

$\implies\:\sf{\dfrac{1}{3}}$

•°• Height of image is 3 cm.

(Size of image is 3 times less than that of the size of the object)

[ Ray diagram is in the same attachment. ]

As clearly observed, the image is diminished or small as compared to the object.